21.

METHODS OF
D I F F E R E N T I AT I O N
A N D A P P L I C AT I O N S
O F D E R I VAT I V E S

1. INTRODUCTION
The rate of change of one dependent quantity with respect to another dependent quantity has great importance.
E.g. the rate of change of displacement of a particle with respect to time is called its velocity and the rate of change
of velocity is called its acceleration. The rate of change of a quantity ‘y’ with respect to another quantity ‘x’ is known
as the derivative or differentiable coefficient of ‘y’ with respect to ‘x.’
According to the first principle of calculus, if y = f(x) is the derivative function, then the derivative of f(x) with respect
to x is given by:
dy f(x+h)-f(x)
f’(x) = = lim
dx h→0 h
Note: y’, y1, Dy can also be used to denote the derivative of y with respect to x. Differentiation is the process of
finding the derivative of a function. ⇒ sin β > 0; cos α < 0

2. DERIVATIVES OF SOME STANDARD FUNCTIONS

Different types of differentiation formulae

d d n
(a) (constant) = 0 (f) (x ) = nxn–1
dx dx

d x d x
(b) (e ) = ex (g) (a ) = ax loge a
dx dx

d 1 d 1
(c) (loge x) = (h) (logax) =
dx x dx xloge a

d d
(d) (sin x) = cos x (i) (cos x) = – sin x
dx dx

d d
(e) (tan x) = sec2x (j) (cot x) = – cosec2x
dx dx
2 1 . 2 | Methods of Differentiation and Applications of Derivatives

d d
(k) (sec x) = sec x tan x (z) (cosec x) = – cosec x cot x
dx dx

d 1 d 1
(l) (sin–1x) = ,–1<x<1 (aa) (cos–1x) = – ,– 1 < x < 1
dx 1−x 2 dx 1 − x2
d 1 d 1
(m) (tan–1x) = ,x∈R (ab) (cot–1x) = – ,∀x∈R
dx 1 + x2 dx 1 + x2
d 1 d −1
(n) (sec–1x) = |x|>1 (ac) (cosec–1x) = | x | >1
dx | x | x2 − 1 dx | x | x2 − 1
d d
(o) (sinh x) = coshx (ad) (coshx) = sinhx
dx dx

d d
(p) (tanh x) = sech2x (ae) (cothx) = –cosech2x
dx dx

d d
(q) (sechx) = – sechx tanhx (af) (cosechx) = –cosechx cothx
dx dx

d 1 d 1
(r) (sinh–1x) = ,∀x∈R (ag) (cosh–1x) = ,|x|>1
dx 1+x 2 dx 2
x −1
d 1 d 1
(s) (tanh–1x) = ,x±1 (ah) (coth–1x) = ,x≠±1
dx 1 − x2 dx 2
x −1
d 1 d −1
(t) (sech–1x) = – ,|x|<1 (ai) (cosech–1x) = ,∀ x ∈ R
dx | x | 1 − x2 dx | x | x2 + 1
d ax
(u) (e sin bx) = eax (a sin bx + b cos bx) = a2 + b2 eaxsin (bx + tan–1 b/a)
dx

d ax
(v) (e cos bx) = eax (a cos bx – b sin bx) = a2 + b2 eaxcos (bx + tan–1 b/a)
dx

d x d 1
(w) |x|= (x ≠ 0) (aj) log | x | = , (x ≠ 0)
dx |x| dx x

d
(x) [x] = 0, ∀ x ∈ –I (where [ . ] denotes greatest integer function)
dx

d
(y) {x} = 1, ∀ x ∈ R (where { . } denotes fractional part function)
dx

PLANCESS CONCEPTS

If the function is continuous, you do not have to apply the first principle method to check differentiability.
You can go directly for dy/dx and check whether dy/dx exists on both the left and right sides and are
equal. If dy/dx does not exist for either one side or both the sides or if both the derivatives exist, but are
not equal or finite, then the function is not differentiable.
E.g. Let y = sin(x) be a continuous function. Check differentiability at x = π/2.On checking for dy/dx =
cos(x) on both the right and left sides, it is found to be equal and finite. Hence, y = sin(x) is differentiable
at x = π/2.
 M a them a ti cs | 21.3

PLANCESS CONCEPTS

Misconception:
(i) In dy/dx, dy or dx does not exist individually.
dy 1 dy dx
(ii) = only if both and exist.
dx dx dx dy
dy
Rohit Kumar (JEE 2012, AIR 79)

3. PRODUCT RULE
d dv du d d(w) d(v) d(u)
(a) (uv) = u +v (b) (uvw) = uv + uw + vw
dx dx dx dx dx dx dx

4. DIVISION RULE
 du   dv 
v   – u 
d u  dx   dx  , where v ≠ 0 (known as the quotient rule)
 =
dx  v  v 2

5. CHAIN RULE
d dy dy dt
(f(g(x))) = f’(g(x)) . g’(x) or = .
dx dx dt dx
d d d
Note: (a) (f(x) ± g(x)) = (f(x)) ± (g(x)), on condition that both f’(x) and g’(x)exist
dx dx dx
d d
(b) (k f(x)) = k (f(x)), where k is any constant
dx dx

x 4 + x2 + 1 dy
Illustration 1:If y = ; = px + q, find p and q. (JEE MAIN)
2
x + x +1 dx

Sol: Differentiate and compare.

y=
(x2 + 1)2 − x2
y=
(( x + 1) + x ) ( x
2 2
+1− x ) = x + 1 – x ⇒ dy = 2x – 1 ⇒ p = 2 and q = –1
2

x2 + x + 1 x2 + x + 1 dx

((x1 + 1) + x)(x2 + 1 − x)
y=
x2 + x + 1

x3 + 2x dy
Illustration 2: If y = , then find . (JEE MAIN)
e x dx

Sol: Differentiate

ex (3x2 + 2x ln2) − (x3 + 2x )ex (3x2 + 2x ln2) − (x3 + 2x )

y’ = =
e2x ex
2 1 . 4 | Methods of Differentiation and Applications of Derivatives

tan–1 x − cot −1 x  dy 
Illustration 3: If y = , find   . (JEE MAIN)
−1
tan x + cot −1
x  dx x =1

Sol: Differentiate and put x = 1.

y=
2
π
(
tan–1 x − cot −1 x )  π
......  tan−1 + cot −1 x =


2
dy 2 2 4  dy  4 2
⇒ = + = ⇒   = =
dx 2 2
π(1 + x ) π(1 + x ) 2
π(1 + x )  dx x =1 2 × π π

Illustration 4: Differentiate the following functions with respect to x: (JEE MAIN)

1 2x
(i) 3x + 2 + (ii) esec + 3cos–1x (iii) log7(log x)
2
2x + 4
1
Sol: (i) Let y = 3x + 2 + = (3x + 2)1/2 + (2x2 + 4)–1/2
2
2x + 4
1 −1
dy 1 −1 d  1 2
−1 d
= (3x + 2) 2 (3x + 2) +  −  (2x + 4) 2 (2x2 + 4)
dx 2 dx  2 dx

1 3
1 − 1 − 3 2x
= (3x + 2) 2 . (3) –   (2x2 + 4) 2 . 4x = –
2 2 2 3x + 2 (2x + 4)3/2
2

2x
(ii) Let y = esec + 3 cos–1x

dy 2 d  1  2  d   1 
= esec x . (sec2x) + 3  −  = esec x .  2sec x (sec x)  + 3  − 
dx dx    dx   
 1 − x2   1 − x2 

2x  1  2  1 
= 2 sec x (sec x tan x) esec + 3−  = 2 sec2 x tan x esec x – 3  + 
   
 1 − x2   1 − x2 

log(logx)
(iii) Let y = log7(log x) = (using change of base formula)
log7

dy 1 d 1 1 d 1
= (log(logx)) = . (log x) =
dx log7 dx log7 logx dx xlog7logx

dy 1
Illustration 5: Find , if y =3 tan x + 5 logax + x – 3ex + . (JEE MAIN)
dx x

Sol: Chain rule.

1
We have y = 3 tan x + 5 logax + x – 3ex +
x
dy d  x 1
Thus, =  3tanx + 5loga x + x − 3e + 
dx dx  x
1
d d d d d 1 5 1 −
= (3 tan x) + (5 loga x) + ( x)– (3ex) +   = 3 sec x + (logea) + x 2 – 3e – x .
2 -1 x –2
dx dx dx dx dx  x  x 2
 M a them a ti cs | 21.5

Illustration 6: Let f, g and h be differentiable functions. If f(0) = 1, g(0) = 2, h(0) = 3 and the derivative pairwise
products at x = 0 are (fg)’ (0) = 6, (gh)’(0) =4 and (hf)’ (0) = 5, then compute the value of (fgh)’ (0). 
 (JEE ADVANCED)
Sol: Product rule
(fgh)’ = f’gh + fhg’ + fgh’  …….(i)
(fg)’ (0) = 6 ⇒ (fg’ + gf’) (0) = 6
(gh)’ (0) = 4 ⇒ (gh’ + hg’) (0) = 4
(hf)’ (0) = 5 ⇒ (hf’ + fh’) (0) = 5
1 1
(fgh)’ = (2f’gh + 2fg’h + 2fgh’) = (f’gh + f’gh + fg’h + fg’h + fgh’ + fgh’)
2 2
1 1
= [h(f’g + fg’) + g(f’h + fh’) + f(g’h + gh’)] = [h(fg)’ + g(fh)’ + f(gh)’]
2 2
1 1
⇒ (fgh)’ (0) = [(3)(6) + (2)(5) + (1)(4)]= [18 + 10 + 4] = 16
2 2

6. TRIGONOMETRIC TRANSFORMATIONS
In case of inverse trigonometric functions, it becomes very easy to differentiate a function by using trigonometric
transformations.Given below are some important results on trigonometric and inverse trigonometric functions.

2 tanx
(a) sin2x = 2 sin x cos x =
1 + tan2 x
1 − tan2 x
(b) cos 2x = 2 cos2 x –1 = 1 – 2 sin2 x = cos2 x – sin2 x =
1 + tan2 x
2 tanx
(c) tan 2x = (n) sin 3x = 3 sin x – 4 sin3x
2
1 – tan x
3tanx − tan3 x
(d) cos 3x = 4 cos3x – 3 cos x (o) tan 3x =
1 – 3tan2 x
π
(e) sin–1 x + cos–1 x = tan–1 x + cot–1 x =sec–1 x + cosec–1 x =
2

(f) sin–1 (–x) = – sin–1 x (p) cos–1 (–x) = π – cos–1 x

(d) tan–1 (–x) = – tan–1 x (q) cosec–1 (–x) = – cosec–1 x

(h) cot–1 (–x) = π – cot–1 x (r) sec–1 (–x) = π – sec–1 x

2 2 2 2
(i) sin–1 x ± sin–1 y = sin–1 (x 1 − y ± y 1 − x ) (s) cos–1 x ± cos–1 y = cos–1 (xy  1 − x 1 − y )

x±y
(j) tan–1 x ± tan–1 y = tan–1 1  xy (t) 2sin–1 x = sin–1 (2x 1 − x2 ) (Be aware of ranges for ‘x’)

 2x   2x   1 − x2 
  –1   –1  
(k) 2cos x = cos (2x – 1)
–1 –1 2
(u) 2 tan x = tan
–1 –1
= sin =cos  2 
 1 − x2   1 + x2  1 + x 
π 1 − x 
(l) – tan–1x = tan–1  1 + x  (v) 3 sin–1x = sin–1 (3x – 4x3)
4  
3x − x3
(m) 3 cos–1x = cos–1 (4x3 – 3x) (w) 3 tan–1x = tan–1
1 − 3x2
2 1 . 6 | Methods of Differentiation and Applications of Derivatives

PLANCESS CONCEPTS

Some useful substitutions in finding derivatives are given below.

Sl. No. Function Substitution

(i) x = a sinθ or a cosθ

a2 – x2
(ii) x = a tan θ or a cotθ
x2 + a2
(iii) x = a sec θ or a cosecθ
x2 – a2
(iv) x = a cos 2θ
a–x
a+ x

(v) x2 = a2 cos 2θ
a2 – x2
a2 + x2

(vi) x = a sin2 θ
ax – x2

(vii) x = a tan2 θ
x
a+ x

(viii) x = a sin2 θ
x
a–x

(ix) x = a sec2 θ– b tan2 θ

(x – a)(x – b)

(x) x = a cos2 θ+ b sin2 θ

(x – a)(b – x)

Rohit Kumar (JEE 2012, AIR 79)

 1 + x2 + 1 
  dy
Illustration 7: If y = cot 
–1
x  , find dx .  (JEE MAIN)
 

Sol: Substitute a suitable trigonometric function in place of x and simplify.

Putting x = tan θ, we have
 sec θ + 1   1 + cos θ  θ 1
y= cot–1  tan θ  = cot–1  sin θ  = cot–1(cot θ/2) = = tan–1x
    2 2
dy 1 1
\ = .
dx 2 1 + x2
 M a them a ti cs | 21.7

PLANCESS CONCEPTS

To differentiate a complex function, put x in some trigonometric form so that the function can be easily
differentiated and then put back x in the form of an inverse trigonometric function.
E.g. Find the derivatives of sec–1 [1/(2x2 – 1)] with respect to 1 − x2 at x = 1/2.
Sol. Putting x = cosθ, we get
1
u = sec–1 = sec–1(sec2θ) = 2θ and y = 1 − x2 = sinθ
2cos2 θ − 1

du 2 2 du
∴ u = 2sin–1y ⇒ = = Thus, =4
dy 1−y 2
x 2 dy x =1/2

Ravi Vooda (JEE 2009, AIR 71)

a−x  dy
Illustration 8: If y = (a − x)(x − b) – (a – b) tan–1  x − b  , find . (JEE ADVANCED)
  dx

Sol: Use Substitution to simplify the given expression and then differentiate.
Let x = a cos2θ + b sin2θ
∴ a – x = a – a cos2θ – b sin2 θ = (a– b) sin2θ … (i)
x – b = a cos θ + b sin θ – b = (a – b) cos θ
2 2 2
… (ii)
∴ y = (a – b) sinθ cosθ – (a – b) tan–1 (tanθ)
(a − b)
y= sin2θ – (a – b) θ
2
 dy 
 
dy dθ (a − b)cos2θ − (a − b) 1 − cos2θ a−x 
Then, =   = = = tan θ=   [From (i) and (ii)]
dx  dx  (b − a)sin2θ sin2θ  x −b
 
 dθ 

7. LOGARITHMIC DIFFERENTIATION
If differentiation of an expression is done after taking log on both the sides, then it is known as logarithmic
differentiation. This method is used when a given expression is in one of the following forms:
(a) (f(x))g(x)
Let y = (f(x))g(x)
Taking logarithm of both the sides, we get log y = g(x) log f(x)
Differentiating with respect to x, we get

1 dy 1 dy  g(x) 
. = g(x) . . f’(x) + log f(x) . g’(x) ⇒ = y f '(x) + logf(x).g'(x) 
y dx f(x) dx  f(x) 

dy  g(x) 
⇒ = (f(x))g(x)  f(x) f '(x) + logf(x).g'(x) 
dx  
2 1 . 8 | Methods of Differentiation and Applications of Derivatives

Short method: The derivative of [f(x)]g(x)can be directly written as:

d  d 
(f(x))g(x) = f(x)g(x)  {g(x)logf(x)} 
dx  dx 
(b) Product of three or more functions
 f '(x) g'(x) h'(x) 
If y = f(x).g(x).h(x), then y’ = f(x).g(x).h(x).  + + 
 f(x) g(x) h(x) 

100
Illustration 9: If f(x) = ∏ (x− n)n(101−n) , find f’(101)/f(101). (JEE ADVANCED)
n=1

Sol: Use logarithms followed by differentiation.

100
f(x) = ∏ (x− n)n(101−n)
n=1

100
f '(x) 100 n(101 − n)
ln f(x) = ∑ n(101 − n)ln(x− n) ⇒ =∑
f(x) n=1 x − n
n=1

100
f '(101) n(101 − n) 100x101
⇒
f(101)
= ∑ = = 5050
n=1 101 − n 2

Illustration 10: Find the derivative of (sin x)cosx. (JEE MAIN)

Sol: Take logarithms on both sides and differentiate.

d d 
(sin x)cos x = (sin x)cos x  dx {cos xlog(sinx)}  = (sin x)cos x [cos x . cot x – sin x . log sin x]
dx  

Illustration 11: Find the derivative of xx with respect to x.  (JEE MAIN)

Sol: Use logarithms to find the derivative.

Let y = xx or y = xx
log y = x log x = ex ln x

1 dy 1 dy d xlnx d
= x   + log x (or) = e = ex ln x dx (x ln x)
y dx x dx dx

dy  1 
= xx(1 + log x) (or) = ex ln x x. + lnx1
dx  x 
d x
∴ = x x x (1 + loge x ) (or) = xx(1 + ln x)
dx
d x
Hence (x ) = xx(1 + ln x)
dx

Illustration 12: Differentiate xsin x with respect to x. (JEE MAIN)

Sol: Similar to the previous illustration.

First method: Let y = xsin x
 M a them a ti cs | 21.9

∴ log y = log xsinx = sinx log x

1 dy sinx
Differentiating we get, = + cos x log x
y dx x

dy  sinx 
∴ =  + cos xlogx  xsinx
dx  x 
Second Method: y = xsinx = esinx log x

dy  sinx   sinx 
Therefore, = esin x log x  x + cos xlogx  =  + cos xlogx  xsinx
dx    x 

cos−1 ( x +1 )
Illustration 13: Differentiate e with respect to x.  (JEE ADVANCED)

Sol: Similar to the previous illustration.

−1 (x +1) dy −1 dy
Let y = ecos Then, = ecos (x +1) . [cos–1(x+1)]
dx dx

−1 (x +1) −1 d −1 −1
= ecos . . (x + 1) = ecos (x +1)
1 − (x + 1)2 dx 1 − (x + 1)2

Illustration 14: Differentiate xsinx, x > 0, with respect to x. (JEE ADVANCED)

Sol: Let y = xsin x

Taking logarithm on both the sides, we get log y = sin xlog x

1 dy d d
Therefore, . = sin x (log x) + logx (sin x)
y dx dx dx

1 dy 1 dy  sinx   sinx 
. = (sin x) + log x cos x = y + cos xlogx  = xsin x  x + cos xlogx 
y dx x dx  x   

Illustration 15: Find f’(x), if f(x) = (sin x)sin x, for all 0 < x <π. (JEE ADVANCED)

Sol: Similar to the previous illustration.

log y = log(sin x)sin x = sin x log (sin x)

1 dy d 1 d
Then, = (sin x log (sin x)) = cos x log (sin x) + sin x . . (sin x)
y dx dx sinx dx

= cos x log (sin x) + cos x = (1 + log (sin x)) cos x

dy
⇒ = y((1 + log (sinx)) cos x) = (1 + log (sinx)) sin xsinx cosx
dx

−1 x
Illustration 16: Differentiate xcos with respect to x. (JEE MAIN)
−1 x
Sol: (i) Let y = xcos
−1 x.ln x
Then, y = ecos
Differentiating both the sides with respect to x, we get
2 1 . 1 0 | Methods of Differentiation and Applications of Derivatives

dy −1 d dy −1  d d 
= ecos x.log x (cos–1x . log x) ⇒ = xcos x logx. (cos−1 x) + cos−1 x. (logx)
dx dx dx  dx dx 
dy −1  − logx cos−1 x 
⇒ = xcos x  + 
dx  2 x 
 1−x
−1 x
(ii) Let (sinx)cos
−1 x.log sin x
Then, y = ecos
Differentiating both the sides with respect to x, we get
dy −1 d
= ecos x.logsin x (cos–1x . log sin x)
dx dx

dy −1  d d 
⇒ = (sinx)cos x cos−1 x. (log sin x) + log sinx (cos–1 x)
dx  dx dx 

dy −1 
 1  −1  
⇒ = (sinx)cos x cos−1 . cos x + log sin x  
dx  sinx  2 
 1−x 

dy −1 
 logsinx 
⇒ = (sinx)cos x cos−1 x . cot x − 
dx  1 − x2 

dy sin x..... ∞
Illustration 17: Find , if y = (sinx)sin x (JEE ADVANCED)
dx 

Sol: Write the given expression as y = (sin x)y and proceed.

We have y = (sin x)y , Therefore, log y = y log sin x
Differentiating both the sides with respect to x, we get

1 dy d dy cos x dy
=y (log sin x) + (log sin x) =y + log sin x
y dx dx dx sinx dx

1  dy dy y 2 cot x y 2 cot x
⇒  − logsinx  = y cot x or = =
y  dx dx 1 − y logsinx 1 − log y

8. DIFFERENTIATION OF IMPLICIT FUNCTION

If in an equation, both x and y occur together, i.e. f(x, y) = 0, and the equation cannot be solved for either x or y,
then x (or y) is called the implicit function of y (or x).
E.g. x3 + y3 + 3axy + c = 0, xy + yx = ab, etc.
Working rule for finding the derivative
First method:
(a) Every term of f(x, y) = 0 should be differentiated with respect to x.
(b) The value of dy/dx should be obtained by rearranging the terms.
Second method:
dy −∂f / ∂x ∂f ∂f
If f(x, y) = constant, then = , where and are the partial differential coefficients of f(x, y) with
dx ∂f / ∂y ∂x ∂y
respect to x and y, respectively.
 M a them a ti cs | 21.11

Note: Partial differential coefficient of f(x, y) with respect to x can be defined as the ordinary differential coefficient
of f(x, y) with respect to x keeping y constant.
∂z ∂x
E.g. z = x2 y ⇒ = x2 , = 2xy
∂y ∂x

sinx dy (1 + y)cos x + y sinx

Illustration 18: If y = , then prove that = . (JEE ADVANCED)
cos x dx 1 + 2y + cos x − sinx
1+
sinx
1+
cos x
1+
1 + ....to ∞

Sol: Write the R.H.S. in terms of x and y. Then differentiate the equation on both sides.
sinx (1 + y)sinx
We have, y = = ⇒ y + y2 + y cos x = (1 + y) sin x
1 + ( (cos x) / (1 + y) ) 1 + y + cos x
On differentiating both the sides with respect to x, we get
dy dy dy dy
+ 2y + cos x – y sin x = sin x + (1 + y) cos x
dx dx dx dx
dy dy (1 + y)cos x + y sinx
⇒ {1 + 2y + cos x – sin x} = (1 + y) cos x + y sin x ⇒ =
dx dx 1 + 2y + cos x − sinx

1
Illustration 19: If f(x) = x + , then compute the value of f (100) . f’(100). (JEE MAIN)
1
2x +
1
2x +
2x + ......∞
1
Sol: Same as above y – x =
1
2x +
1
2x +
2x + ......∞
1
⇒ y–x= ⇒ (y – x) (x + y ) = 1 ⇒ y2 – x2 = 1
2x + y − x

⇒ (f(x))2 = 1 + x2 ⇒ 2(f(x)) × f’(x) = 2x

⇒ f(100) . f’(100) = 100

(lnx )(lnx )∞ dy
(
Illustration 20: If y = (lnx)lnx ) , then find
dx
.(JEE MAIN)

Sol: Same as above

ln y = y ln(lnx)

1 y 1  y  1 − yln(lnx)  y
× y’ = + ln(lnx).y ' ⇒ y’  − ln(lnx)  = ⇒ y’   =
y xlnx  y  xlnx  y  xlnx
y2
⇒ y’ =
(xlnx)(1 − ln(lnx)y)

y dy x+y
Illustration 21: If log (x2 + y2) = 2 tan–1  x  , then prove that = .(JEE ADVANCED)
  dx x−y

Sol: Differentiating both the sides of the given relation with respect to x,
2 1 . 1 2 | Methods of Differentiation and Applications of Derivatives

d d  −1  y  
We get, [log(x2 + y2)] = 2 tan   
dx dx   x  
 dy 
1 d 1 d y 1  dy  x2  x dx − y.1 
⇒ . (x2 + y2) = 2 . .   ⇒ 2x + 2y  = 2.  
x2 + y 2 dx 1 + (y / x)2 dx  x  x2 + y 2  dx  x2 + y 2  x2 
 
 dy   dy  dy dy dy dy x+y
⇒ 2 . x + y  = 2 x − y ⇒ x + y =x –y ⇒ (y – x) = – (x + y) ⇒ =
 dx   dx  dx dx dx dx x−y

9. DIFFERENTIATION OF PARAMETRIC FORM

x and y are sometimes given as functions of a single variable, E.g. x = φ(t) and y = ψ(t) are two functions, where t
is a variable. Then in such cases, x and y are called parametric functions or parametric equations and t is called the
dy
parameter. To find in parametric functions, the relationship between x and y should be obtained by eliminating
dx
the parameter t and then it should be differentiated with respect to x. However, it is not convenient to eliminate the
dy dy dy / dt
parameter every time. Therefore, can also be found by using the formula = .
dx dx dx / dt
dy
E.g. If x = a(θ + sinθ) and y = a(1 – cosθ), then find .
dx
dx dy
Sol: Given that = a(1 + cosθ), = a(sinθ)
dθ dθ
Therefore, x= 0(θ + sin θ). y= a(1 − cos θ)
dy
dy asin θ θ
= dθ = = tan
dx dx a(1 + cos θ) 2
dθ
dy
Note: It may be noted here that can be expressed in terms of the parameter only without directly involving the
main variables x and y. dx

dy
Illustration 22: Find , if x = a cosθ and y = a sinθ. (JEE MAIN)
dx
Sol: Differentiate the two equations w.r.t. θ and eliminate θ.
Given that x = a cos θ and y = a sinθ
dx dy
Therefore, = – a sinθ, = a cosθ.
dθ dθ
dy
dy acos θ
Hence, = dθ = = – cotθ.
dx dx −asin θ
dθ

dy π
Illustration 23: If x = a sec2θ and y = a tan3θ, where q∈ R, find at θ = . (JEE MAIN)
dx 8
Sol: Differentiation of Parametric form.
dy
3 a tan2 θ × sec2 θ
dy
dx
= dθ =
dx
3 π dy
= tanθ; At θ = ,
2asec θ × sec θ tan θ 2 8 dx
=
3
2
( 2 −1 )
dθ
 M a them a ti cs | 21.13

dy
Illustration 24: If x = cosecθ – sinθ and y = cosecnθ – sinnθ, then find . (JEE ADVANCED)
dx

Sol: Differentiation of Parametric form.

x = cosec θ – sinθ
⇒ x2 + 4 = (cosec θ – sin θ)2 + 4 = (cosec θ + sin θ)2 … (i)
and y2 + 4 = (cosecnθ – sinnθ)2+ 4 = (cosecnθ + sinnθ)2 … (ii)
 dy 
 
dy dθ n(cosecn−1 θ)( − cosecθ cot θ) – nsinn−1 θ cos θ
Now, =   =
dx  dx  − cosecθ cot θ − cos θ
 
 dθ 

n(cosecnθ cot θ + sinn−1 θ cos θ ncot θ(cosecnθ + sinn θ) n(cosecnθ + sinn θ)

= = =
(cosecθ cot θ + cos θ) cot θ(cosecθ + sin θ) (cosecθ + sin θ)

dy
Illustration 25: Find if x = at2and y = 2at.  (JEE MAIN)
dx

Sol: Given that x = at2, y = 2at

dx dy
Therefore, = 2at and = 2a.
dt dt
dy
dy 2a 1
Hence, = dt = = .
dx dx 2at t
dt

dy   π 
Illustration 26: If x = cos3t and y = sin3t, then find , for  t ∈  0,   . (JEE MAIN)
dx   2 
dy
dx dy dy 3 sin2 t cos t
Sol: = – 3 cos2t sint (≠ 0) ⇒ = 3 sin2 t cos t ⇒ = dt = = – tan t
dt dt dx dx -3cos2 t sint
dt
dy 16t(1 − t 4 )
Illustration 27: If y = sec 4x and t = tan x, then prove that = . (JEE ADVANCED)
dt (1 − 6t2 + t 4 )2
Sol: Write y in terms of t and differentiate.

( )
2
2
1 1 + tan2 2x 1 + 2t / (1 − t )
y= = =
cos 4x 1 − tan2 2x
( )
2
1 − 2t / (1 − t2 )

(1 − t2 )2 + 4t2 1 + t 4 – 2t2 + 4t2 1 + t 4 + 2t2 dy 16t(1 − t 4 )

y= = = ; =
(1 − t2 )2 − 4t2 1 + t 4 − 2t2 − 4t2 1 + t 4 − 6t2 dt (1 − 6t2 + t 4 )2

2
1 + lnt 3 + 2lnt ydy  dy 
Illustration 28: If x = and y = , then show that = 2x   + 1 (JEE MAIN).
t2 t dx  dx 
Sol: Differentiation of Parametric form.
dy dy / dt
=
dx dx / dt
2 1 . 1 4 | Methods of Differentiation and Applications of Derivatives

dy t ( 0 + 2(1 / t) ) − (3 + 2lnt)1 2 − 3 − 2lnt  1 + 2lnt 

= = =–  
dt t2 t2  t
2


dx t2 ( 0 + (1 / t) ) − (1 + lnt)2t t − 2t − 2tlnt 1 − 2 − 2lnt  1 + 2lnt 

= = = =–  
dt t 4
t 4 t 3
 t
3

2
dy  dy  1 + lnt 2 dy
⇒ =t ⇒ 2x   + 1 = 2. .t + 1 = 3 + 2lnt = yt = y
dx dx
  t 2 dx

10. DIFFERENTIATING WITH RESPECT TOANOTHER FUNCTION

du
Suppose u = f(x) and v = g(x) are two functions of x. To find the derivative of f(x) with respect to g(x), i.e. to find , the
dv
du du / dx
formula = is used. Thus, to find the derivative of f(x) with respect tog(x), both are differentiated with
dv dv / dx
respect to x and then the derivative of f(x) with respect to x is divided by the derivative of g(x) with respect to x. The
procedure is demonstrated in illustration 29.

Illustration 29: Differentiate sin2x with respect to ecosx. (JEE MAIN)

Sol: Differentiate both the functions with respect to the common variable and use parametric form.
du du / dx
Let u(x) = sin2x and v(x) = ecosx. We want to find the value of =
dv dv / dx
du dv
Clearly, = 2 sin x cos x and = ecosx (– sin x) = – (sin x) ecosx
dx dx
du 2sinx cos x 2cos x
Hence, = = − .
dv ( − sinx)ecos x
ecos x

 1 + x2 + 1 − x2 
Illustration 30: Differentiate   with respect to 1 − x 4 . (JEE ADVANCED)
 1 + x2 − 1 − x2 
 

Sol: Similar to the previous illustration.

2
 
 1 + x2 + 1 − x2   1 + x2 + 1 − x2  1 + x2 + 1 − x2 + 2 1 − x 4
Let u =  =   =
 1 + x2 − 1 − x2  (1 + x 2
) – (1 − x 2
) 2x2
 
 
x2  0 +  ( −4x3 ) / (2 1 − x 4 )   −  1 + 1 − x 4  2x
1 + 1 − x4 du     
⇒ u= ⇒ =
x2 dx x4

 ( −2x5 ) / ( 1 − x 4 )  − 2x  1 − x 4 + 1   4  4  
     x +  1 − x  1 + 1 − x4 
du du −2x  
⇒ =     ⇒ =  
dx x4 dx 1 − x4  x4 
 
−2x  x4 + 1 − x4 + 1 − x4  = −2x  1 − x4 + 1  
=     ….(i)
x 4
1−x  4  x 4
1−x  4 

Let v = 1 − x4
 M a them a ti cs | 21.15

dv 1 dv −2x3
= (–4x3) ⇒ =  ….(ii)
dx 2 1−x 4 dx 1 − x4

du du dv du −2x  1 − x4 + 1  1 − x4 du (1 + 1 − x 4 )
= / ⇒ =   ⇒ =
dv dx dx dv 4
x 1−x 4   −2x3 dv x6

11. DIFFERENTIATION OF DETERMINANTS

To differentiate a determinant, one row (or column) at a time should be differentiated, keeping others unchanged,
which is illustrated by the examples given below.

f(x) g(x) d f '(x) g'(x) f(x) g(x)

(i) If F(x) = , then {F(x)} = +
u(x) v(x) dx u(x) v(x) u'(x) v '(x)

d f '(x) g(x) f(x) g'(x)

Also {F(x)} = +
dx u'(x) v(x) u(x) v '(x)
f g h
(ii) If F(x) =  m n Where f, g, h,  , m, n, u, v, w are functions of x and differentiable, then
u v w

f ' g' h' f g h f g h f' g h f g' h f g h'

F’(x) =  m n +  ' m' n' +  m n ⇒ F(x) =  ' m n +  m' n +  m n'
u v w u v w u' v ' w' u' v w u v' w u v w'

sec θ tan2 θ 1
Illustration 31: If f(x) = sec θ tanx x , then find f’(θ). (JEE MAIN)
1 tanx − tan θ 0

Sol: Above discussed method.

0 0 0 sec θ tan2 θ 1 sec θ tan2 θ 1

f’(x) = sec θ tanx x + 0 sec2 x 1 + sec θ tanx x
1 tanx − tan θ 0 1 tanx − tan θ 0 0 sec2 x 0

sec θ tan2 θ 1 sec θ tan2 θ 1

⇒ f’(θ) = 0 sec2 θ 1 + sec θ tan θ θ = (tan2θ – sec2θ) – sec2θ (qsecθ –secθ)= – 1 – (sec3θ) (θ – 1)
1 0 0 0 sec2 θ 0

12. SUCCESSIVE DIFFERENTIATION

dy
If the first derivative of a function y = f(x) is also a differentiable function, then it can be further differentiated
dx
with respect to x. The derivative thus obtained is called the second derivative of y with respect to x and is denoted
d2 y d2 y d3 y
by . If is also differentiable, then its derivative is called the third derivative of y and is denoted by .
dx2 dx2 dx3
dn y
Similarly, denotes the nth derivative of y. This process is known as successive differentiation and all these
dxn
2 1 . 1 6 | Methods of Differentiation and Applications of Derivatives

derivatives are called as successive derivatives of y.

The following symbols are also used to denote the successive derivatives of y = f(x):

y1, y2, y3, ………. Yn, ………..

dy d2 y d3 y dn y d
y’, y’’, y’’’, ………. yn, ……….. ⇒, , , …….. …….., ⇒ Dy, D2y, D3y, ……….Dny ………… (where D ≡ )
dx dx2 dx3 dx n dx
The following symbols are used to denote the value of the nth derivative at x = a.

 dn y 
yn(a), yn(a),   , Dny(a) & fn(a)
 dxn 
  x =a

PLANCESS CONCEPTS

n
dn y  dy 
Misconception: ≠ 
 dx  
n
dx Rohit Kumar (JEE 2012 AIR 79)

13. Nth DERIVAVES OF SOME STANDARD FUNCTIONS

(a) Dn(ax + b)m = m(m – 1) (m –2) ……… (m – n + 1) an(ax + b)m-n
m! m!
(b) If m ∈ N and m > n, thenDn(ax + b)m = an(ax + b)m–n; Dn(xm) = xm–n
(m − n)! (m − n)!
(c) Dn(ax + b)n = n! an; Dn(xn) = n!
 1  ( −1)n n!an 1 ( −1)n n!
(d) Dn  ax + b  = ; Dn  x  =
  (ax + b)n+1   xn+1

( −1)n−1 (n − 1)! ( −1)n−1 (n − 1)!

(e) Dn{log(ax + b)} = an; Dn(log x) =
(ax + b)n xn
(f) Dn(eax) = aneax
(g) Dn(amx) = mn (log a)n amx
 π  π
(h) Dn{sin (ax + b)} = an sin  ax + b + n  ; Dn(sin x) = sin  x + n 
 2   2 
 π  π
(i) Dn {cos (ax + b)} = ancos  ax + b + n  ; Dn (cos x) = cos  x + n 
 2  2

 b
(j) Dn {eax sin (bx + c)} = (a2 + b2)n/2 eax sin  bx + c + ntan−1 
 a

 b
(k) Dn {eax cos (bx + c)} = (a2 + b2)n/2 eax cos  bx + c + ntan−1 
 a
 −1 x ( −1)n−1 (n − 1)!sinn θ sinnθ a
(l) Dn  tan  =
a ,where θ = tan–1  x 
 an  
1
(m) Dn (tan–1 x) = (–1)n–1(n – 1)! sinnθ sin nθ, where θ = tan–1  x 
 
 M a them a ti cs | 21.17

14. LEIBNITZ THEOREM

If u and v are two functions such that their nth derivative exists, then the nth derivative of their product can be found
by the following formula:
Dn(uv) = (Dnu)v + nC1Dn–1u. Dv + nC2Dn–2u. D2v +……+……+nCn-1Du. Dn-1v + ……..+ u.Dnv
The nth derivative of a product of two functions can be found out by using this theorem. While using this theorem,
the second function in the product is the function whose successive derivative starts to vanish (if it is possible) after
some stepsand the first function is a function whose nth derivative is easily known.

Illustration 32: If y = x3 cos x, find Dny. (JEE MAIN)

Sol: Leibnitz theorem

Choose cos x as the first function and x3 as the second function
Dn(cos x, x3) = Dn(cos x) (x3) + nC1Dn–1(cos x) (Dx3) +nC2Dn–2(cos x) . (D2x3) + nC3Dn–3(cos x) . (D3x3)

 nπ   (n − 1)π  (n − 1)  (n − 2)π  n(n − 1)(n − 2)  x + (n − 3)π 

= x3cos  x +  + n.3x cos
2
x + + 6x.cos x + + .6.cos  
 2   2  1.2  2  1.2.3  2 
 nπ   nπ   nπ   nπ 
= x3cos  x +  + 3nx sin  x +
2
 – 3n(n – 1)x cos  x +  – n(n – 1) (n – 2) sin  x + 
 2   2   2   2 

APPLICATION OF DERIVATIVES
1. THE INTERPRETATION OF THE DERIVATIVE
dy
If y = f(x) be a given function, then the derivative/differential coefficient f’(x) or
at the point P(x1, y1) is called the
dx
trigonometric tangent of the angle ψ (say), which the positive direction of the tangent to the curve at P makes with
dy
the positive direction of the x-axis. Therefore, represents the slope of the tangent.
dx
dy Y
Thus, f’(x) = =Y 
dx (x1 ,y1 ) nt
ge
y=f(x)
Then,
Tan

o
90
dy
(a) The inclination of the tangent with x-axis = tan–1 dx P(x1, y1)
No
rm
dy al
(b) Slope of the tangent =
dx 
dx V
(c) Slope of the normal = –
dy
Figure 21.1

2. EQUATION OF TANGENT
 dy 
(a) Equation of tangent to the curve y = f(x) at A(x1, y1) is given by y – y1 =   (x – x1)
 dx (x1 ,y1 )
If the tangent at P (x1, y1) of the curve y = f(x) is parallel to the x-axis (or perpendicular to the y-axis), then
Ψ = 0, i.e its slop will be equal to zero.
2 1 . 1 8 | Methods of Differentiation and Applications of Derivatives

 dy 
⇒ m=   =0
 dx (x1 ,y1 )
The converse also holds true. Thus, the tangent at (x1, y1) is parallel to the x-axis.
 dy 
⇒   =0
 dx (x1 ,y1 )
(b) If the tangent at P (x1, y1) of the curve y = f(x) is parallel to the y-axis (or perpendicular to the x-axis), then Ψ
= π / 2 and its slope will be infinity, i.e.
 dy 
m=   =∞
 dx (x1 ,y1 )
The converse also holds true. Thus, the tangent at (x1, y1) is parallel to the y-axis.
 dy 
⇒   =∞
 dx (x1 ,y1 )
(c) If at any point P (x1, y1) of the curve y = f(x) the tangent makes equal angles with both the axes, then at the
point P, ψ = x / 4 or 3π / 4.Therefore at P, tan Ψ = dy / dx = ± 1.
The converse of the result also holds true. Thus, at (x1, y1), the tangent line makes equal angles with both the axes.
 dy 
⇒   =±1
 dx (x1 ,y1 )
(d) Concept of vertical tangent: y = f(x) has a vertical tangent at the point x = x0 if Y

f(x0 + h) − f(x0 )
Lim = ∞ or – ∞, but not both.
h→0 h

E.g. The functions f(x) = x1/3 and f(x) = sin x both have a vertical tangent at x = 0 X
0 if x < 0
But f(x) = x2/3, f(x) =x and f(x) =  have no vertical tangents at x = 0.
1 if x ≥ 0
Figure 21.2
(e) If a curve passes through the origin, then the equation of the tangent at the
origin can be directly written by equating the lowest degree terms present in the
equation of the curve to zero.
E.g. Y

(i) x2 + y2 + 2gx + 2fy = 0

Equation of tangent is gx + fy = 0
(ii) x3 + y3– 3x2y + 3xy2 + x2 – y2 = 0 X
O
Equation of tangent at the origin is x2 – y2 = 0
(iii) x3 + y3– 3xy = 0
Figure 21.3
Equation of tangent is xy = 0
Note: This concept is valid only if the powers of x and y are natural numbers.
(f) Same line could be the tangent and normal to a given curve at a given point.
E.g. In x3 + y3 – 3xy = 0 (folium of Descartes), the line pair xy = 0 is both the tangent and normal at x = 0.

Some common parametric coordinates on a curve that are useful for differentiation
(a) For x2/3 + y2/3 = a2/3, take parametric coordinates x = a cos3qand y = a sin3θ.
(b) For x + y = a , take x = a cos4θ and y = a sin4θ.
 M a them a ti cs | 21.19

xn yn
(c) + = 1,where x = a (sin θ)2/nand y = b(cos θ)2/n.
an
bn

(d) For c2 (x2 + y2) = x2y2, take x = c sec θ and y = c cosec θ.

(e) For y2 = x3 , take x = t2 and y = t3.

Illustration 33: If the tangent to the curve 2y3 = ax2 + x3 at the point (a, a) cuts off intercepts α and β on the
coordinate axes, where a2 + b2 = 61, the value of | a | is ____. (JEE MAIN)
(A) 16 (B) 28 (C) 30 (D) 31

Sol: (C) Write the equation of the tangent and find the value of α and β in terms of a. Then use a2 + b2 = 61 to find
the value of a.
dy 2ax + 3x2
The slope of the tangent is given by = The value of this slope at (a, a) is 5/6.
dx 6y 2
5 x y
Hence, the equation of tangent is y – a = (x – a) ⇒ + =1
6 −a / 5 a / 6
a a
Thus, the x-intercept α is – , and the y-intercept β is .
5 6
a2 a2
From a2 + b2 = 61, we get + = 61 ⇒ a2 = 25 × 36 ⇒ | a | = 30
25 36

3. EQUATION OF NORMAL
Equation of normal at (x1, y1) to the curve y = f(x) isgiven by the following formula:

−1  dy 
(y – y1) = (x – x1) ⇒ (y – y1)   + (x – x1) = 0
 dy   dx (x1 ,y1 )
 
 dx (x1 ,y1 )

Some facts regarding the normal

 dx 
(a) Slope of the normal drawn at point P (x1,y1) to the curve y = f(x) = –  
 dy (x1 ,y1 )

dx dy
(b) If the normal makes an angle of θ with the positive direction of the x-axis, then – = tanθ or = – cotq
dy dx

dx dy
(c) If the normal is parallel to the x-axis, then = 0 or =∞
dy dx

 dx  dy
(d) If the normal is parallel to the y-axis, then   = ∞ or =0
 dy  dx

 dx   dy 
(e) If the normal is equally inclined from both the axes or cuts equal intercept, then –   = ± 1 or   = ± 1
 dy   dx 

 dy 
x1 + y1  
(f) The length of the perpendicular from the origin to the normal is P’ =  dx 
2
 dy 
1+ 
 dx 
2 1 . 2 0 | Methods of Differentiation and Applications of Derivatives

 dy 
(g) The length of the intercept made by the normal on the x-axis is x1 + y1   and the length of the intercept
 dx 
on the y-axis is y1+ x1  dx  .
 
 dy 

Illustration 34: Find out the distance between the origin and the normal to the curve y = e2x + x2 at the point
whose abscissa is 0. (JEE MAIN)
1 2 3 2
(A) (B) (C) (D)
5 5 5 3
Sol: (B) Write the equation of the normal and find the distance of origin from the normal.
The point on the curve corresponding to x = 0 is (0, 1)
dy dy
= 2e2x + 2x ⇒ =2
dx dx x =0
Therefore, the equation of the normal at the point (0, 1) is
y – 1 = (– 1/2) (x – 0) ⇒ 2y + x – 2 = 0
2
Hence, the distance of the point (0, 0) from this line is .
5

4. LENGTH OF TANGENT, NORMAL, SUBTANGENT AND SUBNORMAL

4.1 Tangent
2
 dy 
y 1+ 
 dx 
PT = MP cosec Ψ = y 1 + cot2 ψ = P
dy
dx 
y
4.2 Subtangent 
T O M G
y
TM = MP cot Ψ = Figure 21.4
(dy / dx)

4.3 Normal
2
 dy 
GP = MP sec Ψ = y 1 + tan2 ψ = y 1 +  
 dx 

4.4 Subnormal

 dy 
MG = MP tan ψ = y  
 dx 

Illustration 35: For the parabola y2 = 16x, the ratio of the length of the subtangent to the abscissa is _____.
(A) 2 : 1 (B) 1 : 1 (C) X : Y (D) X2 : Y  (JEE MAIN)
 M a them a ti cs | 21.21

y
Sol: (A) The length of subtangent is
(dy / dx)

dy dy 8
Differentiating, 2y = 16 Hence, =
dx dx y
dx y 2 16x
Thus, the length of the subtangent is y = = = 2x
dy 8 8
Therefore, the ratio of the length of the subtangent to the abscissa = 2x : x = 2 : 1.

Illustration 36: Find out the length of the normal to the curve x = a(θ + sinθ), y = a (1 – cos θ) at θ = π/2.
 (JEE MAIN)
2
 dy 
Sol: Use differentiation of the Parametric form. Length of the normal = 1+ 
 dx 
 dy 
 
dy  dθ  asin θ θ  dy  π
= = = tan ⇒   = tan   = 1
dx  dx  a(1 + cos θ) 2  dx  4
 
 dθ 
π  π
Moreover, at θ = , y = a  1 − cos  =a
2  2
2
 dy 
Therefore, the required length of the normal = y 1 +   = a 1 + 1 = 2a
 dx 

Illustration 37: The length of the subtangent to the ellipse x = a cos t, y = b sin t at t = π/4 is _____.

(A) A (B) B (C) B/ 2 (D) A/ 2  (JEE MAIN)

Sol: (D) Similar to the previous illustration.

dx dy dy b b
= – a sin t and = b cos t; Therefore, =– cot (π / 4) = –
dt dt dx t= π/ 4
a a

dx π a a
Therefore, the length of the subtangent = y = bsin × – =
dy r = π / 4 4 b 2

5. ANGLE OF INTERSECTION OF TWO CURVES

The angle of intersection between two intersecting curves C1 and C2 is defined C1 T1
as the acute angle between their tangents (T1 and T2 or the normals) at the
point of intersection of the two curves.
m1 − m2 
tan ψ = , where m1 and m2 are the slopes of the tangents T1 and T2 T T2
1 + m1m2 (x1,x2)
at the intersection point (x1, y1)
C2
π
Note: If the two curves intersect orthogonally, i.e at right angle, then φ= .
Hence, the condition will be 2 Figure 21.5

 dy   dy 
  .  =–1
 dx 1  dx 2
2 1 . 2 2 | Methods of Differentiation and Applications of Derivatives

Illustration 38: Which of the following options represents the tangent of the angle at which the curves y = ax and
y = bx(a ≠ b > 0)intersect? (JEE ADVANCED)
logab loga / b logab
(A) (B) (C) (D) None of these
1 + logab 1 + (loga)(logb) 1 + (loga)(logb)

Sol: (B) Differentiate the two curves and use the formula for angle between two lines.
Intersection of the two curves is given by ax = bx, which implies that x = 0. If α is the angle at which the two curves
intersect, then
m1 − m2 ax loga − bx logb (loga / b)
tan α = = = (Putting x = 0)
1 + m1m2 x x
1 + a b (loga)(logb) 1 + (loga)(logb)

6. RATE MEASURE
dy
Whenever a quantity y varies with another quantity x, satisfying the rule y = f(x), then (or f’(x)) represents the
dx
dy
rate of change of y with respect to x and (or f’(a)) represents the rate of change of y with respect to x at x = a.
dx x =a

Illustration 39: The volume of a cube increases at the rate of 9 cm3. How fast does the surface area increase when
the length of an E.g. is 10 cm? (JEE MAIN)

Sol: Rate measurer.

Let x be the length of the side, V be the volume and S be the surface area of the cube.
dV dx dx 3 dS d  3  36
Thus, = 9 cm3/s ⇒ 3x2 dt = 9 cm3/s ⇒ = cm/s ⇒ = (6x2) = 12x   = cm2/s
dt dt x 2 dt dt 2
x  x

dS 
⇒  = 3.6 cm2/s
dt  x =10cms

Illustration 40: A man of height 2 meters walks away from a 5-meter lamppost at a uniform speed of 6 meters per
minute. Find the rate at which the length of his shadow increases. (JEE MAIN)

Sol: Use similarity to establish the relation between the rate at which length of shadow increases and speed of the
man.
B
Let AB be the lamp-post. Let at any time t, the man CD be at a distance x metres from the
lamp-post and y metres be the length of his shadow CE.
D
dx
Then, = 6 meters / minute [given] … (i)
dt
Clearly, the triangles ABE and CDE are similar X y
AB AE 5 x+y A C E
⇒ = ⇒ = ⇒3y = 2x
CD CE 2 y Figure 21.6
dy dx dy dy
⇒ 3 =2 ⇒3 = 2(6) [Using (i)]⇒ = 4 meters / minute
dt dt dt dt

Illustration 41: An object has been moving in the clockwise direction along the unit circle x2 + y2 = 1. As it passes
through the point (1/2, 3 /2), its y-coordinate decreases at the rate of 3 units per second. The rate at which the
x-coordinate changes at this point is ______ units per second.

(A) 2 (B) 3 3 (C) 3 (D) 2 3  (JEE MAIN)

 M a them a ti cs | 21.23

Sol: (B) Differentiate and proceed.

dx 1 3 dy
We find that when x = and y = given that = – 3 units/s and x2 + y2 = 1.
dt 2 2 dt
dx dy
Differentiating x2 + y2 =1, we get 2x + 2y =0
dt dt
1 dy 1 dx 3 dx
Putting x = ,y= 3 /2 and = – 3, we get + (– 3) = 0 ⇒ = 3 3 (increasing)
2 dt 2 dt 2 dt

Illustration 42: A given right circular cone has a volume p.The largest right circular cylinder that can be inscribed
in the cone has a volume q. The ratio of p to q is _____. (JEE MAIN)
(A) 9 : 4 (B) 8 : 3 (C) 7 : 2 (D) None of these

Sol: (A) Let H be the height of the cone and α is its semi-vertical angle.
Let x be the radius of the inscribed cylinder and h be its height.
h = QL = OL – OQ = H – x cot a
1
p= π(H tan α)2 H … (i)
3
V = volume of the cylinder = px2 (H – x cot α)  O
dV
= π(2Hx – 3x2 cot α) 
dx x
dV
Hence, =0 ⇒x=0 Q
dx
2 d2 V
x= H tan α, = – 2πH < 0, so
3 dx=2
2  M P
x Htan α
3
2 4 1 4 Figure 21.7
V is maximum when x = H tan α and q = Vmax = π H2 tan2α H = p [using (i)]
3 9 3 9
Therefore, p : q = 9 : 4

7. APPROXIMATION USING DIFFERENTIALS

To calculate the approximate value of a function, differentials may be used, wherein the differential of a function is
equal to its derivative multiplied by the differential of the independent variable.
dy = f’(x)dx or df(x) = f’(x) dx

PLANCESS CONCEPTS

For the independent variable ‘x’, increment Dx and differential dx can be made equal, but the same
cannot be applied in case of the dependent variable ‘y,’ i.e. Dy ≠ dy.
Therefore, the approximate value of y when the increment ∆ x is given to the independent variable x in
y = f(x) is
dy
y + Dy = f(x + Dx) = f(x) + . Dx
dx
⇒ f(x + Dx) = f(x) + f’(x) Dx
Vaibhav Gupta (JEE 2009 AIR 54)
2 1 . 2 4 | Methods of Differentiation and Applications of Derivatives

Illustration 43: Find the approximate value of the square root of 25.2. (JEE MAIN)

Sol: Consider a function f(x) = x and differentiate to get the derivative. Then replace x by x+Dx and proceed.
1
Let f(x) = x , so f’(x) =We can write 25.2 as 25 + 0.2
2 x
By taking x = 25 and Dx = 0.2, now f(x + Dx) = f(x) + f’(x). Dx
1 1
⇒ x + .Dx = 25 + . 0.2
2 x 2 25
0.2
= 5+ = 5 + 0.02 = 5.02
10

Illustration 44: What is the approximate change in the volume V of a cube of side x meters caused by increasing
the side by 2%? (JEE MAIN)
dV
Sol: Differentiate the equation V = x3 and use the relation DV = Dx.
dx
Let ∆(x) be the change in x and DV be the corresponding change in V
∆x
Given that × 100 = 2
x
dV
We know that V = x3 ∴ = 3x2
dx
dV 2x
Therefore, DV = Dx ⇒ DV= 3x2Dx = 3x2 × = 0.06 x3m3
dx 100
The approximate change in volume is 0.06 x3m3.

Illustration 45: What is the approximate value of cos 40º? (JEE ANDANCED)
(A) 0.7688 (B) 0.7071 (C) 0.7117 (D) 0.7

Sol: (A) Take a function f(x) = cos x and proceed.

π π π π
Let f(x) = cos x. 40º = 45º – 5º = – ×5= – radians
4 180 4 36
A differential is used to estimate the change in cos x
π π  π 
When x decreases from to –  
4 4  36 

f’(x) = – sin x and df (x)= f’(x) h = – h sin x

π π
With x = and h = – , df is given by
4 36

 π  π π 1 π 2
df = – f’(x)h = –  −  sin   = . = = 0.0617
 36  4 36 2 72

cos 40 ≡ cos 45 + 0.0617 ≡ 0.7071 + 0.0617 = 0.7688.

8. SHORTEST DISTANCE BETWEEN TWO CURVES

It has been found that the shortest distance between two non-intersecting curves is always along the common
normal (wherever defined).
 M a them a ti cs | 21.25

Illustration 46: Find out the shortest distance between the line y = x – 2 and the parabola y = x2 + 3x + 2.
 (JEE MAIN)

y = x + 3x + 3y y
2
Sol: The distance would be minimum at the point on the parabola where
the slope of the tangent is equal to the slope of the given line. 
y=x-2
Let P(x1, y1) is the point closest to the line y = x – 2 P=(x1, y1)
x’ x
dy O
Then, = slope of the line
dx (x
1 ,y1 )

⇒ 2x1+ 3 = 1 ⇒ x1 = –1 and y1 = 0
y’
Therefore, point (–1, 0) is the closest and its perpendicular distance from the
line y = x – 2 gives the shortest distance. Figure 21.8

3
⇒ Shortest distance = units
2

Illustration 47: Which of the following points of the curve y = x2 is closest to (4, – ½)? (JEE MAIN)
(A) (1, 1) (B) (2, 4) (C) (2/3, 4/9) (D) (4/3, 16/9)

Sol:(A) Using distance formula find the distance of the given point from the curve and find the minima.
Let the required point be (x, y) on the curve.

Hence, d = (x − 4)2 + (y + 1 / 2)2 should be minimum, which is enough to consider.

D = (x – 4)2 + (y + 1/2)2 = (x – 4)2 + (x2 + 1/2)2
D’ = 4x3 + 4x – 8
Now for critical points
D’ = 0 so x3 + x – 2 = 0 ⇒ x = 1
Clearly D” at x = 1 is 16 > 0.
Thus, D is minimum when x = 1. Hence the required point is (1, 1).

PROBLEM-SOLVING TACTICS

•• Reduce any fractions to be as basic as possible.

•• Recognise when we can use the chain rule. it enables us to differentiate functions that often seem impossible
to differentiate. Whenever you see a nested function, try to assess if the chain rule is needed (it usually is).
•• We always want to start a long chain of differentiation by differentiating the last part of the function to touch
the input - in short, the outermost part of the function.
2 1 . 2 6 | Methods of Differentiation and Applications of Derivatives

FORMULAE SHEET

dc d du
=0 (cu) = c
dx dx dx

d du dv d dv du
(u ± v) = ± (uv) = u +v
dx dx dx dx dx dx

du dv
v −u dy dy du
d u dx dx =
  = dx du dx
dx  v  v2

d n d n du
x = nxn–1 u = nun–1 dx
dx dx

d x d u du
a = (ln a) ax a = (ln a) au dx
dx dx

d x d u du
e = ex e = eu dx
dx dx

d 1 d 1 du
logax = logau =
dx (lna)x dx (lna)u dx

d 1 d 1 du
ln x = ln u =
dx x dx u dx

d d du
sin x = cos x sin u = cos u
dx dx dx

d d du
cos x = – sin x cos u = – sin u
dx dx dx

d d du
tan x = sec2 x tan u = sec2 u
dx dx dx

d d du
cot x = – cosec2 x cot u = – cosec 2 u
dx dx dx

d d du
sec x = sec x tan x sec u = sec u tan u
dx dx dx

d d du
cosec x = – cosec x cot x cosec u = – cosec u cot u
dx dx dx

d 1 d 1 du
sin–1 x = sin–1 u =
dx 1−x 2 dx 1−u 2 dx

d 1 d 1 du
tan–1 x = tan–1 u =
dx 1 + x2 dx 1 + u2 dx
 M a them a ti cs | 21.27

 dy 
* Equation of tangent to the curve y = f(x) at A(x1, y1) is y – y1 =   (x – x1)
 dx (x1 ,y1 )

−1
* Equation of normal at (x1, y1) to the curve y = f(x) is (y – y1) = (x – x1)
 dy 
 
 dx (x1 ,y1 )

* Length of Tangent, Normal, Subtangent and Subnormal

2
 dy 
y 1+ 
 dx 
Tangent: PT = MP cosec Ψ = y 1 + cot2 ψ = P
dy
dx 
y
y
Subtangent: TM = MP cot Ψ = 
(dy / dx)
T O M R G
2
 dy  Figure 21.9
Normal: GP = MP sec Ψ = y 1 + tan2 ψ = y 1 +  
 dx 

 dy 
Subnormal: MG = MP tan Ψ = y   C1 T1
 dx 
* Angle of Intersection of Two Curves

m1 − m2 T T2
tan Ψ = ,  (x1,x2)
1 + m1m2
C2
where m1 and m2 are the slopes of the tangents T1 and T2 at the intersection
point (x1, y1). Figure 21.10

Solved Examples

JEE Main/Boards f’(0–) =

lim f(x) − f(0) lim −( −x) − 0
= x →0 = –1
x →0 – x −x
Example 1: Show that the function f(x) = | x | is Since, f’(0+)≠f’(0–), the function is not differentiable at
continuous at x = 0, but not differentiable at x = 0. x=0

Sol: Evaluate f’(0+) and f’(0-).

Example 2: Find the derivative of the function f(x),
 x, x ≥ 0 defined by f(x) = sin x by 1stprinciple.
We have f(x) = 
 –x, x < 0
Sol: Use the first principle to find the derivative of the
Since lim f(x) = lim f(x) = 0 = f(0) given function.
x →0 + x →0 –
The function is continuous at x = 0 Let dy be the increment in y corresponding to an
increment dx in x. We have
We also have
y = sin x
f(x) − f(0) x−0
f’(0+) = lim = lim =1
x →0 + x x →0 x y +dy = sin (x + dx)
2 1 . 2 8 | Methods of Differentiation and Applications of Derivatives

Subtracting, we get x
y log x = x – y ⇒ y =
dy = sin(x + dx) – sin x = 2 cos (x + dx/2) sin (dx / 2) 1 + logx

Dividing by dx, we obtain dy (1 + logx).1 − x(1 / x) log x

∴ = =
dx (1 + logx)2
(1 + log x)2
δy cos(x + δx / 2)sin(δx / 2)
=
δx (δx / 2)
Taking limits on both side, we get Example 6: If y = cot–1 x2 − 1 + sec–1x, then dy/dx
equals
dy δy sin(δx / 2)
= lim = lim cos(x + dx/2) lim
dx δx →0 δx δx →0 δx →0 ( δx / 2) Sol: Use substitution to simplify the terms and then
differentiate.
= cos x . 1 = cos x
2
Hence, we have d/dx(sin x) = cos x Put x = secθ ; cot-1 x − 1 = cot sec2 θ − 1

π π
Example 3: The derivative of log | x | is = cot–1(tanθ) = – θ = – sec–1 x
2 2
Sol: Use the definition of the modulus to expand the π  π dy
∴ y =  − sec –1 x  + sec–1 x = ⇒ =0
given function. Then evaluate L.H.D. and R.H.D. at the  2  2 dx
critical point.
dy
Let y = log | x | then Example 7: If x2ey + 2xyex + 13 = 0, then equals -
dx
 logx, whenx > 0 Sol: Use the formula for derivative of implicit function.
y= 
log( −x), whenx < 0 Using partial derivatives, we have

dy 1 ∂f
∴ = when x > 0 dy ∂x
dx x ⇒ =
dx ∂f
dy 1 1
and = (– 1) = when x < 0 ∂y
dx −x x
dy 1 dy 2xey + 2yex + 2xyex
⇒ = when x ≠ 0 =–
dx x dx x2ey + 2xex

1+x dy 2xey − x + 2y + 2xy 2xey − x + 2y (x + 1)

Example 4: If y = , then equals =– =–
1−x dx
x2ey − x + 2x x(xey − x + 2)

Sol: Differentiate using u/v rule.

d  −1  cos x  
Differentiating w.r.t., we get Example 8: tan    equals -
dx   1 + sinx  
dy 1 1 − x (1 − x)1 − (1 + x)( −1) cos x
= Sol: Convert in terms of tan and proceed.
dx 2 1+x (1 − x)2 1 + sinx

⇒
1−x 1
⇒
1 1 cos x sin ( π / 2 − x )
1 + x (1 − x)2  =
1 + x (1 − x)3/2 1 + sinx 1 + cos( π / 2 − x)

1  1  π x  cos x 
⇒   = tan  −  ∴ tan–1  1 + sinx 
1−x 2 1 − x   4 2  

  π x  π x 1
Example 5: If xy = ex-y, then dy/dx equals - = tan–1  tan  4 − 2   = − ⇒ Derivative = –
   4 2 2
Sol: Take logarithms on both sides and differentiate.
Taking log on both sides, we get sec x − tanx dy
Example 9: If y = , then equals
sec x + tanx dt
 M a them a ti cs | 21.29

Sol: Simplify the R.H.S. and differentiate. dy

sec x − tanx sec x − tanx dy dx 1
y= . ∴ = =(2x2 + 1)
dz dz
(1 / (2x )
2
sec x + tanx sec x − tanx 2
+ 1) −1
y =(sec x – tan x)2 dx

dy  −4x  2  dy 
∴ = 2(sec x – tan x)(sec x tan x – sec2 x)  2  1 + 3x ∴   =0
dx  2x + 1  3  dz x = −1
3
⇒ – 2 sec x (sec x – tan x)2
  
d 2 −1 1 + x 
1 dy Example 14: Find sin  cot  =
Example 10: If y = , then equals dx 1 − x  
  
(t + 2)(t + 1) dx

Sol: Use the partial fraction method to find the Sol: Use Substitution to simplify the inside the square
derivative of given fn root and then differentiate.

1 1 1  −1 1 + x

y= = – Let y = sin2  cot 
1 − x  .
(t + 2)(t + 1) t + 1 t + 2 
dy 1 1
⇒ =– + Put x = cos 2θ.
dx (t + 1)2
(t + 2)2
 1 + cos2θ  
∴ y = sin2cot–1  1 − cos2θ   = sin2cot–1 (cot θ)
1 1   
Example 11: If x = θ – and y = θ + ,
θ θ
dy 1 − cos2θ 1 − x 1 x
then =? ∴ y = sin2θ= = = –
dx 2 2 2 2
1 dx 1 dy 1
Sol: x = θ – ⇒ =1+ ∴ =–
θ dθ θ2 dx 2
1 dy 1
y=θ+ ⇒ =1– Example 15: Find the equation of the normal to the
θ dθ θ2
π
dy curve y = x + sin x cos x at x = .
dy 1 − (1 / θ2 ) θ − (1 / θ) x 2
∴ = dθ = = = Sol: Find a point on the curve slope of the normal at
dx dx 1 + (1 / θ ) θ + (1 / θ)
2 y
dθ that point.
2 π π π π π
Example 12: Derivative of sin–1 x w.r.t. cos–1 1 − x is - x= ⇒ y = +0 = , so the given point =  ,  .
2 2 2 2 2
Sol: Substitute sin θ in place of x. dy
Now from the given equation = 1 + cos2x – sin2 x
2 dx
Let y = sin–1x and z = cos–1 1 − x
 dy 
⇒  =1+0–1=0
Put x = sin θ⇒ z = cos (cos θ)=θ  dx  π , π 
–1

2 2
dy
∴ y = z and =1 ∴ required equation of the normal is
dz
 1  π −1  π π
y– = x −  ⇒ x – = 0 ⇒ 2x = p
Example 13: Derivative of sec  2
–1  w.r.t. 1 + 3x 2 0  2  2
 2x + 1 
−1
at x = is
3 Example 16: Find the point on the curve y = x2 – 3x at
which tangent is parallel to x-axis.
Sol: Differentiate the two functions and divide.
Sol: Differentiate the given equation and put it equal to
 1  zero and proceed.
Let y= sec–1  2  and z = 1 + 3x
 2x + 1  Let the point at which tangent is parallel to x-axis be
P(x1, y1)
2 1 . 3 0 | Methods of Differentiation and Applications of Derivatives

3 dy 9x
Then it must be on curve i.e.,y1 = x1 – 3x1 We have =–
dx by
dy For curves to intersect at right angles, we must have at
Also differentiating w.r.t. x, we get, = 3x2 – 3
dx the point of intersection.
 dy  3  −9x 
⇒  = 3 x12 – 3 … (i)
 = – 1 ⇒ 27 x = by .
2
dx
 (x1 ,y1 ) 
y  by 
since, the tangent is parallel to x-axis Thus we must have
 dy 
∴  = 0⇒ 3 x12 – 3 = 0 9x2 + by2 = 16 ⇒ 9x2 + 27x – 16 = 0 … (ii)
dx
 (x1 ,y1 )
(i) and (ii) must be identical so 27 = 6b ⇒ b = 9/2.
⇒ x1 = ±1  … (ii)
From (1) and (2); y1 = x12 – 3x1 Example 19: If the tangent at (1, 1) on y2 = x(2 – x)2
When x1 = 1 when x1 = – 1 meets the curve again at P, then P is

y1 = 1 – 3 = – 2, y1 = –1 + 3 = 2 Sol: Solve the equation of the tangent with the equation

∴ points at which tangent is parallel to x-axis are of the curve.
(1, – 2) and (–1, 2). dy
2y = (2 – x)2 – 2x (2 – x)
dx
Example 17: Find the equation of normal to the curve dy 1
= 3x2 – 8x + 4. So = −
x + y = xy, where it cuts x-axis. dx (1,1) 2

Sol: Given curve is x + y = xy  … (i) An equation of tangent at (1, 1) is Y – 1 = (–1/2) (X – 1).

at x-axis y = 0, i.e. Y = (–1/2)x + 3/2. The intersection of this line with
the given curve is given by ((– x/2) + 3/2)2 = x(2 – x)2
∴x+0=x ⇒x=1 0

⇒ x2 – 6x + 9 = 16x + 4x3 – 16x2. So,

∴ Point is A(1, 0)
4x3 – 17x2 + 22x – 9 = 0
Now to differentiation x + y = xy
⇒ (x – 1) (4x – 9) (x + 1) = 0
take log on both sides
Thus x = 1, 9/4, –1. But x = – 1 cannot lie on the given
⇒ log (x + y) = y log x
curve so required point is (9/4, 3/8).
1  dy  1 dy
∴ 1 +  = y . +(log x)
x+y  dx  x dx
JEE Advanced/Boards
 dy 
Putting x = 1, y = 0, 1 +  =0
 dx  Example 1: Examine differentiability of f(x) at
 dy 
⇒  =–1 ∴ slope of normal = 1 1 − cos x
 dx (1,0)  , x≠0
y −0 x = 0 for f(x)=  x sinx
Equation of normal is, =1 ⇒ y=x–1  1
x −1 , x=0
 2
Example 18: If the curve y2 = 6x, 9x2 + by2 = 16, cut Sol: Find the left and right hand derivative of the
each other at right angles then the value of b is function f(x) about the point x = 0.
dy First we obtain L.f ‘(0)
Sol: Equate the product of from the two equations
to -1. dx
 f( −h) − f(0)   1  1 − cosh 1 
= lim   = lim  − h   hsinh − 2 
The intersection of the two curves is given by 9x + 6bx 2
h→0  −h  h→0   
= 16  … (i)
dy 3  hsinh + 2(1 − cosh)   0 
Differentiating y2 = 6x, We have =
dx y = lim   ;  in from 
h→0  2
2h sinh   0 
Differentiating 9x + 6y = 16
2 2 
 M a them a ti cs | 21.31

  h3 h5   h2 h4 h6   ( −1)n h   π
hh − + − .....  − 2  − + − .....   = lim   = (–1)n–1Rf’  nπ + 
  3! 5! 

 2! 4! 6!



h→0
 −h   2
= lim  
h→0  h3 
 2h2  h − + .....    f (nx + ( π / 2) + h) − f (nx + ( π / 2) ) 
  3!   = lim  
    h→0 h
 
  1 1   1 2   
 h4  −  +  −  h2   { } { } 
 sin−1  cos (nx + ( π / 2) + h) − sin−1 cos (nπ + ( π / 2))
 
  12 3!   5! 6!    = lim  
= lim    h→0
 h 
h→0
 3
 h2    
2h  h − + ..... 



 3! 


 { } {
 sin−1 ( −1)n cos (( π / 2) + h) − sin−1 ( −1)n cos( π / 2)

= lim 
} 

  1 1   1 2  
h→0
 h 
 h  −  +  −  h + .....    
  12 3!   5! 6!    = 0 and
= lim     sin−1 {( −1)n+1 sinh}   sin−1 {sin( −1)n+1 h} 
h→0
  h2   = lim   = lim  
2 1 − + .....  h→0 h h→0 h
  3!      
   
( −1)n+1 sin−1 sin h ( −1)n+1 h
 f(0 + h) − f(0)  = lim = lim = (– 1)n+1
Rf’(0) = lim   h→0 h h→0 h
h→0
 h 
(Which is equal to (– 1)n–1)
 1  1 − cosh 1  
= lim   −   = 0,
h→0
 h  hsinh 2    π  π
Thus we find Lf’  nx +  = Rf’  nx + 
 2  2
similarly as above i.e. L.f’(0) = Rf’(0)
⇒ f(x) is differentiable at x = 0 ∴ f(x) is differentiable at (nx + π/2)

dy
Example 2: Examine differentiability of the function f(x) Example 3:If x (1 + y) + y (1 + x) = 0, then equals
dx
π Sol: Simplify the equation given and then differentiate
= sin–1 (cos x) at x = np + . where n ∈ I.
2 it.
Sol: Similar to the previous example. We have
 π
first, we obtain Lf’  nx +  x (1 + y) + y (1 + x) = 0  … (i)
 2 
 f (nx + ( π / 2) − h) − f (nx + ( π / 2) )  ⇒ x (1 + y) = –y (1 + x)
= lim  
h→0  −h 
  On squaring both sides x2(1+y) = y2(1+x)
⇒ x2 – y2 + x2y – xy2 = 0
{ } {
 sin−1 cos (nx + ( π / 2) − h) − sin−1 cos (nx + ( π / 2))} 
= lim   ⇒ (x – y) (x + y + xy) = 0
h→0  −h 
  x – y ≠ 0 [For y = x does not satisfy (1)]


= lim 
{ } {
 sin−1 ( −1)n cos ( ( π / 2) − h) − sin−1 ( −1)n cos( π / 2)
}  ∴ x + y + xy = 0⇒ y = –
x
(1 + x)
h→0 −h 
  dy 
 (1 + x).1 − x.1 
 1
∴ =–   =– .
{ } (1 + x)2
2
 sin−1 sin( −1)n h − sin−1 0  dx  (1 + x)
 

= lim  
h→0  −h 
  dy
Example 4: If xyyx = 1, then equals –
dx
( −1)n sin−1 sin h − sin−1 0
= lim Sol: Use logarithms on both sides and then differentiate
h→0 −h
Taking log on both sides, we have y log x + x log y = 0
2 1 . 3 2 | Methods of Differentiation and Applications of Derivatives

Now using partial derivatives, we have Sol: Differentiation w.r.t another function.
dy y / x + log y y(y + xlog y) Putting x = tanθ
=– ⇒–
dx logx + x / y x(x + y logx)  x   tan θ 
y = sin–1   = sin–1 
 =q
 1+x
2
  sec θ 
dy
Example 5: If 1 − x2 + 1 − y 2 = a(x – y) then
equals – dx  1 − x2   1 − tan2 θ 
& z = cos–1  2
 = cos–1 


 1 + tan2 θ  = 2q
1 + x   
Sol: Use substitution for x and y.
Putting x = a sin A, y = a sin B, then given relation ∴ Required derivative = ½
becomes dy
dy dθ θ 1
cos A + cos B = a(sin A – sin B) ⇒ = ==
dz dz 2θ 2
 A +B  A −B  dθ
⇒ 2a cos   cos  
 2   2 

 A +B  A −B 
Example 8: If y = sin–1 ( )
sinx then
dy
dx
equals–
= 2a cos   sin  
 2   2  Sol: Differentiation of function
 A +B
.Divided
Divide
… andmultiply
multiplybby
y cos   dy 1 1
 2  = . .cosx
dx 1 − sinx 2 sinx
 A −B    A +B 
⇒ cot   = a  cos   ≠ 0 1 + sinx 1
 2    2   = = 1 + cosecx
2 sinx 2
⇒ A – B = 2 cot–1a ⇒ sin–1 x – sin–1y = 2 cot–1a

1 1 dy dy 1 − y2 v µ dv
– =0 ⇒ = Example 9: If + = 6 then =?
1−x 2
1−y 2 dx dx 1−x 2 µ v dµ

Sol: Square the given equation and proceed.

1 1
Example 6: If x2 + y2 = t – , x4 + y4 = t2 + then
t t 2
v µ v µ
dy + = 6 ⇒ + +2 = 36
equals µ v µ v
dx
Sol: Eliminate t from the first and the second equation ⇒m2 + v2 = 34 mv
and then find the derivative.
Differentiating both sides w.r.t. µ we have
Squaring the first equation, we have
dv dv
1 2µ + 2ν = 34 ν + 34 µ
x + y + 2x y = t +
4 4 2 2 2
–2 dµ dµ
t2
1 1 dv dv µ − 17ν
⇒ t2 + + 2x2y2 = t2 + – 2 (from second equation) ⇒ 2[17µ – n] = 2 [µ –17 n] ∴ =
dµ dµ 17µ − v
t2 t2
1
⇒ x2y2 = – 1⇒ y2 = –
 2 
dy
x2 Example 10: If x = exp. tan–1  y − x  , then equals
dy 2 dy 1  x2  dx
∴ 2y = ⇒ = 3  
dx x 3 dx x y
Sol: Simplify the given equation and differentiate.
Example 7: The derivation of Taking log on both sides, we get
 x  1 − x 2   y − x2 
sin 
–1
2
 w.r.t. cos–1 
  1 + x2
 is
 log x = tan  2 
–1
 1+x     x 
 M a them a ti cs | 21.33

⇒ tan(log x) = (y – x2) / x2 ⇒ y = x2 + x2 tan (log x) p’(x) = 2yy’ ⇒ p’’(x) = 2yy’’ + 2y’2 ⇒ p’’’(x) = 2yy’’’ + 4y’ y’’
dy  3 d2 y 
∴ = 2x + 2x tan (log x) + x sec2(log x) d d 3
dx Also 2 y  =2 (y y’’)
dx  dx2  dx
 
⇒ 2x [1 + tan (log x)] + x sec2(log x)
= 2[y3y’’’ + 3y’2y’y’’ ] = y2 [2yy’’’ + 6 y’y’’] = p(x) p’’’(x)
 4x3 
d
Example 11: Find cos–1  27 − x  Example 14: If the tangent at the point P(at2, at3) on
dx  
the curve ay2 = x3 intersects the curve again at the point
 4x3    x 3  x   Q, find the point Q.
 − x  –1  4   − 3  
Sol: Let y = cos  27 –1
 = cos   3 
    3   Sol: Solve the equation of the tangent and the equation
of the curve.
x x
= cosθ⇒θ = cos–1  3  ay2 = x3⇒ 2 ay dy/dx = 3x2
3  
 3x2  3a2 t 4 3
∴ y = cos (4 cos θ – 3 cos θ) = cos (cos 3θ) = 3q
–1 3 –1 Slope of tangent at P is  =  = t
 2ay  2 3 2
 p 2a t
x
∴ y = 3cos–1  3  Let Q be (at12, at13). Slope of line
 
at13 − at3 t12 + tt1 + t2
dy −1 1 −3 PQ = =
∴ =3 = . at13 − at2 t1 + t
dx 1 − (x / 9) 3
2
9−x 2
which must be the slope of tangent at P. Hence,
 x2 − y 2  t12 + tt1 + t2 3t
dy
Example 12: If cos  2 
2  = log a then
–1
= = ⇒ 2 t12 – tt1 – t2 = 0
 x + y  dx t1 + t 2
t
Sol: Take cosine on both sides and then apply ⇒ (t1 – t) (2t1 + t) = 0 ⇒ t1 = –
2
componendo and dividendo.
 at2 at3 
 x2 − y 2  Thus, Q has coordinates  ,− 
 4 8 
cos  2
–1 
2  = log a 
x +y 
x2 − y 2 Example 15: Show that the curves ax2 + by2 = 1 and cx2
⇒ = cos (log a) = k (say)
x2 + y 2 1 1 1 1
+ dy2 = 1 cut orthogonally if, – = –
a b c d
by componendo and dividends,
dy
Sol: Equate the product of from the two equations
2 2
(x − y ) + (x + y ) 2 2
k +1 to -1. dx
∴ =
2 2
(x − y ) − (x + y ) 2 2 k −1
Let the two curves cut each other at the point (x1, y1);
then
2x2 k +1 x k +1
⇒ 2
= ∴ = ax12 + by12 = 1 … (i)
−2y k −1 y k −1
& cx12 + dy12 = 1 … (ii)
Differentiating both sides w.r.t. ‘x’ we get
From (i) and (ii), we get
1 x dy dy y
– 2 =0 ⇒ =
y y dx dx x =(a – c) x12 + (b – d) y12 = 0 … (iii)
Slope of the tangent to the curve
Example 13: If y2 = p(x) is a polynomial of degree 3,  dy  ax
ax2 + by2 = 1, at (x1, y1) is given by,   =– 1
d  3d y 2  dx (x1 ,y1 ) by1
then 2 y  is equal to Slope of the tangent to the curve
dx  dx2 
 
cx2 + dy2 = 1, at (x1, y1) is given by,
Sol: Find first order, second order and third order
derivative of p(x).
2 1 . 3 4 | Methods of Differentiation and Applications of Derivatives

 dy  cx dy
=– 1 Differentiating both sides w.r.t. x, then = 2x2 + x
  dx
 dx (x1 ,y1 ) dy1
 Tangents make equal angles with coordinate axes.
If the two curves cut orthogonally, we must have,
 ax1   cx1  dy
∴ = ± 1 or 2x2 + x = ± 1 or
 −   −  dx
 by1   dy1 
2x2 + x + 1 ≠ 0 and 2x2 + x – 1 = 0
⇒ acx12 + bdy12 = 0  … (iv)
or 2x2 + 2x – x – 1 = 0
From (iii) and (iv), we have
(If 2x2 + x + 1 = 0 then x is imaginary)
a−c b−d 1 1 1 1
= ⇒ – = – 1
ac bd a b c d or (2x – 1) (x + 1) ∴ x = ,–1
2
1 2 1 1 1 5
Example 16: Find the acute angle between the curves y From (1), x= , y= . + . =
= | x2 – 1 | and y = | x2 – 3 | at their points of intersection 2 3 8 2 4 24
when x > 0. 2 1 1
and for x = –1, y = – + =–
3 2 6
Sol: Solve the two curves and find the slope for the 1 5   1
two tangents. Proceed to find the angle between the hence point are  ,  and  −1, − 
two lines.  2 24   6

For the intersection of the given curves Example 18: The side of the rectangle of the greatest
| x2 – 1 | – | x2 – 3 | ⇒ (x2 – 1)2 = (x2 – 3)2 area, that can be inscribed in the ellipse x2 + 2y2 = 8,
are given by
⇒ (x2 – 1)2 – (x2 – 3)2 = 0
(A) 4 2 , 4 (B) 4, 2 2
⇒ [(x2 – 1) – (x2 – 3)] [(x2 – 1) + (x2 – 3)] = 0
(C) 2, 2 (D) 2 2 , 2
⇒ [2x2 – 4] = 0 ⇒ 2x2 = 4 ⇒ x = ± 2
Sol: (B) Consider a point on the ellipse and write the
neglecting x = – 2 as x > 0 expression for the area of the rectangle. Then find the
We have point of intersection as x = – 2 maximum area using first and second order derivative.

Here y = | x – 1 | = (x – 1) in the neighbouring of

2 2 x2 y 2
Any point on the ellipse + = 1 is
8 4
x= 2 and y = – (x2 – 3) in the neighbouring of x = 2
(2 2 cos θ, 2 sin θ) [see figure]
 dy   dy 
⇒   = 2x = 2 2 and   = – 2x = – 2 2 y
dx
 c1  dx c1
(22 cos, 2 sin)
Hence, if θ is angle between them,

4 2 0 x
2 2 − ( −2 2) 4 2
⇒ tan θ = = =  
1 + 2 2( −2 2) −7  7 
 
4 2
∴ θ = tan–1  7  A = area of the inscribed rectangle
 
= 4(2 2 cos θ) (2 sin θ) = 8 2 sin2q
dA π
Example 17: At what points on the curve = 16 2 cos 2θ = 0⇒θ =
dθ 4
2 1
y = x3 + x2, then tangent make equals angles with d2 A π
3 2 Also = – 32 2 sin2θ< 0 for θ =
coordinate axes. dθ 2 2
Hence, the inscribed rectangle is of largest area if the
Sol.: Find dy/dx and equate it to ± 1. π π
sides are 4 2 cos and 4 sin   i.e. 4 and 2 2 .
2 3 1 2 4 4
Given curve is y = x + x … (i)
3 2
 M a them a ti cs | 21.35

JEE Main/Boards

Exercise 1  1+x + 1−x

Q.18 y = sin–1  
 2 
Methods of Differentiation
 2 2
Q.19 y = sin–1 2ax 1 − a x 
x +3
Q.1 Find the derivative of e , with respect to x.

Q.2 Differentiate, sin(log x), with the respect to x. Q.20 y = a+ a+ x

 ax − b 
Q.3 If x = sinθ, y = – tanθ, find dy/dx. Q.21 y = tan–1  a + bx 
 

Q.4 Differentiate, cos–1 ( x ) , with the respect to x. Q.22 y = log  x + x2 + a2 

 
−1 x
Q.5 Differentiate, em tan , with the respect to x.
Q.23 y = log  sin 1 + x2 
 
Q.6 Differentiate, sin {log(x3 – 1)}, with the respect to x.

Q.7 Differentiate, cos x, with the respect to ex. Application of Derivatives

Q.8 Differentiate the following w.r.t., x : log2(sin x). Q.1 Find the point on the curve y = x2 – 4x + 5, where
tangent to the curve is parallel to x-axis.

Q.9 Differentiate the following w.r.t., x : y = 5log(sin x).

Q.2 If two curves cut orthogonally, then what can we
dy say about the angle between tangents at the point of
Q.10 Find ,when x + y =5 at (4, 9). intersection of the curves.
dx

1 + x  Q.3 Find the slopes of tangent and normal to the curve

Q.11 y = cot–1  1 − x 
  1
f(x)=3x2 –5 at x = .
2
1 + tanx
Q.12 y = Q.4 If the tangent of the curve y = f(x)
1 − tanx
at point (x, y) on the curve is parallel to y– axis, then
dy
what is the value of .
Q.13 y = sin  cos x  dx
 
Q.5 Find a point on the curve y = 2x2 – 6x – 4 at which
 x + a the tangent is parallel to x-axis.
Q.14 y = tan–1  

 1 − ax 
Q.6 Find a point on the curve y = x2 – 4x – 32 at which
−1 x the tangent is parallel to x-axis.
Q.15 y = (sinx)cos

Q.7 Find the equations of tangent and normal to the

Q.16 y cos−1 ( 2 cos x + 3 sinx ) 13 
= curve y = 3
5 − x at (–3, 2).
 

 1 + x2  Q.8 Find equations of tangent to the curve y = 4x − 3 ,

Q.17 y = sec–1   if parallel to x-axis.
 1 − x2 
 
2 1 . 3 6 | Methods of Differentiation and Applications of Derivatives

Q.9 Verify that the point (1, 1) is a point of intersection Q.23 Find the points on the curve y = x3 – 2x2 – 2x at
of the curves x2 =y and x3 +6y= 7 and show that these which the tangent lines are parallel to the line y = 2x – 3.
curves cut orthogonally at this point.
Q.24 Find the angle between the parabolas y2 = 4ax
Q.10 Find the equation of tangent to the parabola y2 = and x2 = 4by at their point of intersection other than
8x which is parallel to line 4x – y + 3 = 0. the origin.

Q.11 Find the equation of tangent to the curve y = – 5x2 Exercise 2

 1 35 
+ 6x + 7 at the point  ,  .
2 4  Methods of Differentiation
Q.12 Find the equation of tangent to the curve xy = c2 Single Correct Choice Type
c 
at the point  ,ck  on it.
k   3x + 4  dy
Q.1 If y = f   & f’(x) = tanx then
2
=
 5x + 6  dx
Q.13 Prove that the tangents to the curve y = x3 + 6 at
the points (–1, 5) and (1, 7) are parallel. (A) tan x3
2
 3x + 4  1
Q.14 At what point on the curve y = x2 does the tangent (B) – 2 tan   .
make an angle of 45º with x-axis?  5x + 6  (5x + 6 )2

x2 y2  3tanx2 + 4 
Q.15 Find the point (s) on the curve + =1 (C) f   tan x2
9 4  5 tanx2 + 6 
parallel to y-axis.  
(D) None
1
Q.16 Find the slope of the normal to the curve x =
y= 2t at t = 2. t x10
Q.2 Let g is the inverse function of f & f’(x) = .
2
If g(2) = a then g’(2) is equal to (1 + x )
Q.17 Show that equation of the tangent to the curve 5 1 + a2 a10 1 + a10
x2 y 2 x x y y (A) (B) (C) (D)
+ =1 at (x0, y0) is 0 + 0 =1. 210 a10 1 + a2 a2
2
a b 2
a2 b2
1 1
Q.3 If y = +
n−m p −m m−n
Q.18 Find the equation of the normal lines to the curve y 1+x +x 1+x + xp −n
= 4x3– 3x + 5 which are parallel to the line 9y + x + 3 = 0.
1
+
m−p
1+x + xn−p
Q.19 Find the equation of normal line to the curve y(x –
2) (x – 3) – x = 7 = 0 at the point where it meets x-axis. dy np
Then at em is equal to :
dx
Q.20 Find the equation of tangent to the curve x +
(A) emnp (B) emn/p (C) enp/m (D) None
y = a at the point (x1, y1) and show that the sum of
its intercepts on axis is constant. Q.4 Let f is differentiable in (0, 6) & f’(4) = 5 then lim
x →2
f(4) − f(x2 )
=
Q.21 Find the equation of the normals to the curve 2−x
3x2 + y2= 8 parallel to the line x +3y= 4. (A) 5 (B) 5/4 (C) 10 (D) 20

Q.22 Find the equation of the tangents to the curve

Q.5 Let  = Lim xm(lnx)n where m, n∈N then
 a2 a2  x →0
x + y = a at the point  ,  .
 4 4 (A)  is independent of m and n
(B)  is independent of m and depend on m
 M a them a ti cs | 21.37

(C)  is independent of n and depend on m 7 

(C) (2, ∞) (D)  , ∞ 
(D)  is dependent on both m and n 2 

Q.13 Let f(x) = sin x; g(x) = x2& h(x) = loge x & f(x) =
cos x x 1
f '(x) df(x)
Q.6 Let f(x) = 2sinx x2 2x then alim = h[g(f(x))] then is equal to :
→0 x dx2
tanx x 1
(A) 2 cosec3x (B) 2 cost (x2) – 4x2 cosec2 (x2)
(A) 2 (B) – 2 (C) – 1 (D) 1 (C) 2x cot x2 (D) – 2cosec2 x

cos x sinx cos x Q.14 Let f(x) = xn, n being a non-negative integer. The
π
Q.7 Let f(x) cos2x sin2x 2cos2x then f’   = b
cos3x sin3x 3cos3x 2 number of value of n for which f’ (p + q) = f’
ab + 2ay
(p) + f’(q) is valid for all p, q > 0 is :
(A) 0 (B) – 12 (C) 4 (D) 1
(A) 0 (B) 1 (C) 2 (D) None of these
 2x 
Q.8 If y = sin–1  2  , then dy/dx at x = π/2 is 100
1 + x  f(101)
Q.15 If f(x) = ∏ (x − n)n(101−n) ; then f '(101)
=
−8 4 n=1
(A) (B)
2 2
π +4 π +4 1 1
(A) 5050 (B) (C) 10010 (D)
8 5050 10010
(C) (D) Does not exists
π2 + 4
 3x2 + 2x − 1 1
Q.9 If f(4) = g(4) = 2; f’(4) = 9 ; g’(4) = 6 then  2 for x ≠
Q.16 Let f(x)=  6x − 5x + 1 3 then f’  1 
 
f(x) − g(x)  1 3
lim is equal to :  −4 for x =
a→ 4  3
x −2
3 (A) is equal to – 9 (B) is equal to – 27
(A) 3 2 (B) (C) 0 (D) None
2 (C) is equal to 27 (D) does not exist
d2 x
Q.10 If y = x + ex then is : Q.17 Let f(x) be a quadratic expression which is positive
dy 2
for all real x. If g(x) = f(x) + f’(x) + f”(x), then for any real
ex
(A) ex (B) – x, which one is correct.
(1 + ex )3
(A) g(x) < 0 (B) g(x) > 0 (C) g(x) = 0 (D) g(x) ≥ 0
ex −1
(C) – (D)
(1 + ex )2 (1 + ex )3 x4 + 4 dy 
Q.18 If y = then  is :
2
x − 2x + 2 dx  x =1/2
Q.11 If f is twice differentiable such that f”(x) = – f(x),
f’(x) = g(x) h’(x) = [f(x)]2 + [g(x)]2 and h(0) = 2, h(1) =4,
(A) 3 (B) – 1 (C) 4 (D) None
then the equation y = h(x) represents :
(A) A curve of degree 2
Q.19 A function f, defined for all positive real numbers,
(B) A curve passing through the origin satisfies the equation f(x2) = x3 for every x > 0. Then the
(C) A straight line with slope 2 value of f’(4)

(D) A straight line with y intercept equal to– 2 (A) 12 (B) 3 (C) 3/2 (D) Cannot be determined

Q.12 Let f(x) = x + 3 ln(x – 2) & g(x) = x + 5 n(x – 1), Q.20 If x = sin t and y = sin 3t, then the value of ‘K’ for
then the set of x satisfying the inequality f’(x) < g’(x) is d2 y dy
which (1 – x2) –x + Ky = 0 is
 7  dx 2 dx
 7
(A)  2,  (B) (1, 2) ∪  − , ∞  (A) 3 (B) 6 (C) 12 (D) 9
 2  2 
2 1 . 3 8 | Methods of Differentiation and Applications of Derivatives

Q.21 If x = lnt & y = t2 – 1 then y”(1) at t = 1 is Q.7 The subnormal at any point on the curve xyn = an+1
is constant for :
(A) 2 (B) 4 (C) 3 (D) None
(A) n = 0 (B) n =1
(C) n = –2 (D) No value of n
Application of Derivatives
Q.8 Equation of the line through the point (1/2, 2) and
Single Correct Choice Type −x2
tangent to the parabola y = +2 and secant to the
2
Q.1 The angle at which the curve y = kekx intersects the curve y= 4 − x2 is
y-axis is (A) 2x + 2y – 5 = 0 (B) 2x + 2y – 3 = 0
(A) tan–1k2 (B) cot–1(k2) (C) y – 2 = 0 (D) None of these
 4 
(C) sec–1  1 + k  (D) None
Q.9 Two curves C1: y = x2 – 3 and C2 : y = kx2, k ∈ R
intersect each other at two different point. The tangent
Q.2 The angle between the tangent lines to the graph
x drawn to C2 at one of the point of intersection A ≡ (a,
y1), (a > 0) meets C1 again at B(1, y2) (y1≠ y2). The value
of the function f(x)= ∫ (2t − 5) dt at the point where the of ‘a’ is
2
graph cuts the x-axis is -
(A) 4 (B) 3 (C) 2 (D) 1
(A) π/6 (B) π/4 (C) π/3 (D) π/2
Q.10 Number of roots of the equation x2.e2–|x| = 1 is:
Q.3 If a variable tangent to the curve x y = c makes 2 3
(A) 2 (B) 4 (C) 6 (D) Zero
intercepts a, b on x and y axis respectively then the
value of a2b is
Q.11 The x-intercept of the tangent at any arbitrary
4 327 3 4
(A) 27c3 (B) c c (C) (D) c3 a b
27 4 9 point of the curve + = 1 is proportional to
2
 π x y2
x sin for x > 0
Q.4 Consider the function f(x) =  x (A) Square of the abscissa of the point of tangency
 0 for x = 0
 (B) Square root of the abscissa of the point of tangency
then the number of points in (0, 1) where the derivative
(C) Cube of the abscissa of the point of tangency
(D) Cube root of the abscissa of the point of tangency
f’(x) vanishes, is
(A) 0 (B) 1 (C) 2 (D) infinite
Q.12 The line which is parallel to x-axis and crosses the
π
curve y= x at an angle of is
Q.5 The tangent to the graph of the function y = f(x) 4
at the point with abscissa x = a forms with the x-axis (A) y = –1/2 (B) x = ½ (C) y = 1/4 (D) y = ½
an angle of π/3 and at the point with abscissa x = b
at an angle of π/4, then the value of the integral, Q.13 The lines tangent to the curves y3 – x2y + 5y – 2x
h
= 0 and x4 – x3y2 + 5x + 2y = 0 at the origin intersect at
∫ f '(x).f "(x)dx is equal to
an angle θ equal to
a
(A) 1 (B) 0 (C) – 3 (D) – 1
(A) π/6 (B) π/4 (C) π/3 (D) π/2
[assume f”(x) to be continuous]
x
 1
Q.6 Let C be the curve y = x (where x takes all real
3
Q.14 Consider f(x) = ∫  t + t  dt and g(x) = 'f‘ for
1  0
values). The tangent at A meets the curve again at B. If x ∈  ,3
the gradient at B is K times the gradient at A then K is 2 
equal to
(A) 4 (B) 2 (C) – 2 (D) ¼
 M a them a ti cs | 21.39

If P is a point on the curve y = g(x) such that the Q.5 The equation of the common tangent to the curves
tangent to this curve at P is parallel to a chord joining y2= 8x and xy = – 1 is (2002)
 1  1  (A) 3y = 9x + 2 (B) y = 2x + 1
the points  , g    and (3, g(3)) of the curve, then
 2  2 
(C) 2y = x + 8 (D) y = x + 2
the coordinates of the point P
 7 65 
(A) can’t be found out (B)  ,  Q.6 Tangents are drawn to the ellipse x2 + 2y2 = 2, then
 4 28  the locus of the mid point of the intercept made by the
 3 5  tangents between the coordinate axes is  (2004)
(C) (1, 2) (D)  , 
 2 6 1 1 1 1
  (A) + =1 (B) + =1
2 2 2
2x 4y 4x 2y 2
Q.15 The co-ordinates of the point on the curve 9y2 = x3
where the normal to the curve makes equal intercepts x2 y 2 x2 y 2
(C) + =1 (D) + =1
with the axes is 2 4 4 2
6 2 6 
 1
(A)  1, 
 3
(
(B) 3, 3 )  8
(C)  4, 
 3
(D)  ,


5 5 5 

Q.7 The angle between the tangent drawn from the
point (1, 4) to the parabola y2 = 4x is  (2004)
π π π π
(A) (B) (C) (D)
Previous Years’ Questions 6 4 3 2

Q.1 The normal to the curve x = a (cos θ + θ sin θ), y = Q.8 The tangent at (1, 7) to the curve x2 – y – 6 touches
a (sin θ – θ cos θ) at any / point "θ” is such that (1983) the circle x2 + y2 + 16 x + 12y + c = 0 at (2005)
(A) It makes a constant angle with the x-axis (A) (6, 7) (B) (– 6, 7) (C) (6, –7) (D) (–6, – 7)
(B) It passes through the origin
Q.9 The tangent to the curve y = ex drawn at the point
(C) It is at a constant distance from the origin
(c, ec) intersects the line joining the point (c – 1, ec–1) and
(D) None of the above (c + 1, ec+1)  (2007)
(A) On the left of x = c (B) On the right of x = c
Q.2 The slope of tangent to a curve y = f(x) at [x, f(x)] is
2x + 1. If the curve passes through the point (1, 2),then (C) At no point (D) At all points
the area bounded by the curve, the x-axis and the line
x = 1 is  (1995)   1 
(x − 1)sin   , if x ≠ 1
(A) 5/6 (B) 6/5 (C) 1/6 (D) 6 Q.10 Let f(x) =   x −1 .
 0, if x = 1

Q.3If the normal to the curve y = f(x) at the point (3, 4)
Then which one of the following is true ? (2008)
3π
makes an angle with the positive x-axis, then f’(3) is
4 (A) f is neither differentiable at x = 0 not at x = 1
equal to (2000) (B) f is differentiable at x = 0 and at x = 1
(A) – 1 (B) –3/4 (C) 4/3 (D) 1 (C) f is differentiable at x = 0 but not at x = 1
(D) f is differentiable at x = 1 but not at x = 0
Q.4 The point(s) on the curve y3 + 3x2 = 12y where the
tangent is vertical, is (are)  (2002)
Q.11 The area of the plane region bounded by the
 4   11 
(A)  ± , −2  (B)  ± ,0  curves x + 2y 2 =
0 and x + 3y 2 =
1 is equal to  (2008)
 3   3 
 5 1 2 4
(A) (B) (C) (D)
 4  3 3 3 3
(C) (0, 0) (D)  ± ,2 
 3 
2 1 . 4 0 | Methods of Differentiation and Applications of Derivatives

JEE Advanced/Boards

Exercise 1 [1 + (dy / dx)2 ]3/2

Q.10 Show that R = can be reduced
to the form d2 y / dx2
Methods of Differentiation
1 1
R2/3 = 2 +
Q.1 Let y = x sin kx. Find the possible value of k for (d y dx2 )2/3 (d2 x dy 2 )2/3
d2 y
which the differential equation +y= 2k cos kx holds Q.11 Suppose f and g are two functions such that f, g:
dx2
true for all x ∈ R. R → R, f(x) = ln(x+ 1 + x2 ) then find the value of xeg(x)
'
  1 
Q.2 Find a polynomial function f(x) such that f(2x) =  f    + g’(x) at x = 1.
f(x) f”(x).   x 
x+y
Q.12 Let f(x) be a derivative function at x = 0 & f  
Q.3 Let f and g be two real-valued differentiable  k 
function on R If f’(x) = g(x) and g’(x) = f(x) “x ∈ R and f(x) + f(y)
= (k ∈ R, k ≠ 0, 2). Show that f(x) is either a
k
( )
f(3) = 5, f’(3) = 4 then find the value of f 2 ( π) − g2 ( π) . zero or an odd linear function.
2
d y
Q.4 Find the value of the expression y3 2 on the (x − a)4 (x − a)3 1
dx
ellipse 3x2 + 4y2 = 12. Q.13 If f(x) = (x − b) 4
(x − b) 3
1 then
4 3
(x − c) (x − c) 1
Q.5 The function f : R → R satisfies f(x2) . f”(x) = f’(x) .
f’(x2) for all real x. Given that f(1) = 1 and f”’(1) = 8, (x − a)4 (x − a)2 1
compute the value of f’(1) + f”(1). 4 2
f’(x) = λ . (x − b) (x − b) 1 .
1 1
− 4 2
Q.6 If 2x = y5 + y 5 then (x2 – 1) (x − c) (x − c) 1
d2 y dy
+x = ky, then find the value of ‘k’. Find the value of λ.
dx 2 dx
Q.14 Let P(x) be a polynomial of degree 4 such that
Q.7 If the dependent variable y is changed to ‘z’ by the P(1) = P(3) = P(5) = P’(7) = 0. If the real number x
substitution y = tan z then the differential equation ≠ 1, 3, 5 is such that P (x) = 0 can be expressed as
2 x = p/q where ‘p’ and ‘q’ are relatively prime, then find
d2 y 2(1 + y)  dy  d2 z (p + q).
=1+ is changed to = cos2z + k
2  dx  2
dx2 1+y   dx
2
 dz 
  , then find the value of k. Q.15 If f : R → R is a function such that
 dx  f(x)=x3+x2f’(1)+xf”(2)+f”’(3) for all x ∈ R, then prove
 x that f(2) = f(1) – f(0).
Q.8 Show that the substitution z = ln  tan  changes
the equation  2 cos(x + x2 ) sin(x + x2 ) − cos(x + x2 )

d2 y dy sin(x − x2 ) cos(x − x2 ) sin(x − x2 )

+ cot x + 4 y cosec2x = 0 Q.16 If f(x) =
dx 2 dx sin2x 0 sin2x2
 d2 y  then find f’(x)
to   + 4y = 0
 dz 2 
 
a+ x b+ x c+ x
sinx Q.17 Let f(x) =  + x m + x n + x . Show that f”(x) = 0
Q.9 Let f(x) = if x ≠ 0 and f(0) = 1. Define the
x
p+ x q+ x r + x
function f’(x) for all x and find f”(0) if it exist.
 M a them a ti cs | 21.41

and that f(x) = f(0) + kx where k denotes the sum of all The function g(x) is defined by g(x) = eax + f(x) ∀ x ∈ R,
the co-factors of the elements in f(0) where ‘a’ is any constant. If g’(0) + g”(0) = 0. Find the
value(s) of ‘a’.
1 1
Q.18 If y = tan-1 2
+ tan–1 2
+ tan–1 Application of derivatives
x + x +1 x + 3x + 3
1 1
2 + tan–1 2 = + ….. to n terms. Q.1 Find the equations of the tangents drawn to the
x + 5x + 7 x + 7x + 13
curves y2 – 2x3 – 4y + 8 = 0 from the point (1, 2).
Find dy/dx, expressing your answer in 2 terms.
7
Q.2 The tangent to y = ax2 + bx + at (1,2) is parallel
2
Q.19 If Y = sX and Z = tX, where all the letter denotes the to the normal at the point (– 2, 2) on the curvey = x2 +
functions of x and suffixes denotes the differentiation 6x +10. Find the value of a and b.
X Y Z s1 t1 Q.3 Find the point of intersection of the tangents
w.r.t. then prove that X1 Y1 Z1 = X3 s t2
2 drawn to the curve x2y = 1 – y at the points where it is
X2 Y2 Z2 intersected by the curve xy = 1 – y .

u
Q.20 If y = tan–1 & Q.4 Find the equation of the normal to the curve y =
1 − u2 (1+ x)y + sin–1 (sin2x) at x = 0.
1  1  1 
x = sec–1 2 ,u ∈  0,  ∪  ,1  Q.5 A function is defined parametrically by the equation
2u − 1  2  2 
dy  2 1
prove that 2 +1=0 2t + t sin if t ≠ 0
dx f(t) = x =  t and
 0 if t = 0

x  1−x 
Q.21 If y = tan–1 + sin  2 tan−1 , 1
1+ 1−x  2
1 + x   sint
2
if t ≠ 0
 g(t) = y =  t
dy  0
The find for x ∈ (–1, 1)  if t = 0
dx
Find the equation of the tangent and normal at the
1 + sinx + 1 − sinx point for t = 0 is exist.
Q.22 If y = cot–1 ,
1 + sinx − 1 − sinx
41x3
dy  π π  Q.6 A line is tangent to the curve f(x)= at the point
find if x ∈  0,  ∪  , π  . 3
dx  2 2  P in the first quadrant, and has a slope of 2009. This line
intersects the y-axis at(0, b). Find the value of ‘b’.
Q.23 Prove that the second order derivative of a
single valued function parametrically represented by x Q.7 Find all the tangents to the curve y = cos (x + y), –
= φ(t) and y = Ψ(t), α< t <β where φ(t) and Ψ(t) are 2π≤ x ≤ 2π, that are parallel to the line x + 2y = 0
differentiable functions and φ’(t) ≠ 0 is given
 dx   d2 y   d2 x   dy  Q.8 There is a point (p, q) on the graph of f(x) = x2 and
   2  −  2   
d2 y  dt   dt   dt   dt  −8
by = a point (r, s) on the graph of g(x) = where p > 0 and
2 3
dx  dx  x
  r > 0. If the line through (p,q) and (r, s) is also tangent to
 dt 
both the curves at these points respectively, then find
Q.24 (a) If y = y(x) and it follows the relation exy + y cos the value of (p + r)
x = 2, then find (i) y’(0) and (ii) y”(0).
(b) A twice differentiable function f(x) is defined for all Q.9 (i) Use differentials to approximate the values of; (a)
real numbers and satisfies the following conditions f(0) 3
36.6 and (b) 26 .
= 2; f’(0) = –5 and f”(0) = 3
2 1 . 4 2 | Methods of Differentiation and Applications of Derivatives

(ii) If the radius of a sphere is measured as 9 cm with Q.18 Let the function f : [– 4, 4] → [–1, 1] be defined
an error of 0.03 cm, then find the approximate error in implicitly by the equation x=5y – y5=0.
calculating its volume.
Find the area of triangle formed by tangent and normal
to f(x) at x = 0 and the line y = 5.
Q.10 The chord of the parabola y=– a2x2 +5ax – 4
 1
1 Q.19 The normal at the point P  2,  on the curve
touches the curve y = at the point x = 2 and is  2
1−x xy = 1, meets the curve again at Q. If m is the slope of
bisected by that point. Find ‘a’.
the curve at Q, then find | m |.
Q.11 Tangent at a point P1 [other than (0, 0)] on the
curve y = x3 meets the curve again at P2. The tangent Q.20 Let C be the curve f(x) = ln2x + 2lnx and A(a, f(a),
at P2 meets the curve at P3& so on. Show that the B(b, (f(b)) where (a < b) are the points of tangency of
abscissae of P1, P2, P3 ……. Pn, form a GP. Also find the two tangents drawn from origin to the curve C.
area(P1P2P3 ) (i) Find the value of the product ab.
ratio .
area(P2P3P4 ) (ii) Find the number of values of x satisfying the equation
5x f’(x) – x ln 10 – 10 = 0.
Q.12 Determine a differentiable function y = f(x) which
1
satisfies f’(x) = [f(x)]2 and f(0) = – . Find also the Q.21 A particle moves along the curve 6y = x3 + 2. Find
2
the points on the curve at which the y coordinate is
equation of the tangent at the point where the curve
changing 8 times as fast as the x coordinate
crosses the y-axis.

Q.22 A man 1.5 m tall walks away from a lamp post 4.5
Q.13 The curve y = ax3 + bx2 + cx + 5, touches the
m high at the rate of 4 km/hr.
x-axis at P(–2, 0) & cuts the y-axis at a point Q where its
gradient is 3. Find a, b, c (i) How fast is the father end of the shadow moving on
the pavement?
Q.14 Find the gradient of the line passing through the (ii) How fast is his shadow lengthening?
point (2, 8) and touching the curve y = x2.
Q.23 A water tank has the shape of a right circular cone
Q.15 Let f : {0, ∞) → R be a continuous, strictly increasing with its vertex down. Its altitude is 10 cm and the radius
x
2 of the base is 15 cm. Water leaks out of the bottom at a
function such thatf3(x) = ∫ tf
(t)dt . If a normal is drawn
constant rate of 1 cu. cm/sec. Water is poured into the
0 −1 tank at a constant rate of C cu. cm/sec. Compute C so
to the curve y=f(x) with gradient , then find the
2 that the water level will be rising at the rate of 4 cm/sec
intercept made by it on the y-axis. at the instant when the water is 2 cm deep.

Q.16 The graph of a certain function f contains the point Q.24 Water is dripping out from a conical funnel of
(0, 2) and has the property that for each number’p the π
line tangent to y = f(x) at (p, f(p)) intersect the x-axis at semi vertical angle , at the uniform rate of 2 cm3/sec
4
p + 2. Find f(x)
through a tiny hole at the vertex at the bottom. When
the slant height of the water is 4 cm, find the rate of
Q.17 (a) Find the value of n so that the subnormal at decrease of the slant height of the water.
any point on the curve xyn = an+1 may be constant(b)
Show that in the curve y = a.ln (x2 – a2)sum of the length
Q.25 Sand is pouring from a pipe at the rate of 12 cc/
of tangent & subtangent varies as the product of the
sec. The falling sand forms a cone on the ground in
coordinates of the point of contact
such a way that the height of the cone is always 1/6th
(c) If the two curve C1 : x = y2 and C2 : xy = k cut at right of the radius of the base. How fast is the height of the
angles find the value of k. sand cone increasing when the height is 4 cm?
 M a them a ti cs | 21.43

Q.26 A circular ink blot grows at the rate of 2 cm2 per Q.2 The function f(x) = ex + x, being differentiable and
second. Find the rate at which the radius is increasing one to one to one, has a differentiable inverse f–1(x). The
6 22 d –1
after 2 seconds. Use π = value of (f ) at the point f(ln2) is
11 7 dx
1 1 1
Q.27 A variable ∆ABC in the xy plane has its orthocentre (A) (B) (C) (D) None
at vertex ‘B’, a fixed vertex ‘A’ at the origin and the third ln2 3 4
7x2
vertex ‘C’ restricted to lie on the parabola y = 1 + . Q.3 f ’(x) = g(x) and g’(x) = – f(x) for all real x and f(5) = 2 = f
36
‘(5) then f2(10) + g2(10) is-
The point B starts at the point (0, 1) at time t = 0 and
moves upward along the y axis at a constant velocity of (A) 2 (B) 4 (C) 8 (D) None
2 cm/sec. How fast is the area of the triangle increasing
7
when t = sec. Q.4 Differential coefficient of
2 1 1 1
  +m  n−   m+n   −m  n+   m−n
Q.28 At time t > 0, the volume of a sphere is increasing  x m−n  .  x n−   .  x  −m  w.r.t. is
at a rate proportional to the reciprocal of its radius. At      
     
t = 0, the radius of the sphere is 1 unit and at t = 15 the
radius is 2 units. (A) 1 (B) 0 (C) –1 (D) xlmn

(a) Find the radius of the sphere as a function of time t x)

Q.5 Let f(x)=(xx)x and g(x)= x(x then:
(b) At what time t will the volume of the sphere be 27
times its volume at t = 0 (A) f’(1) = 1 and g’(1) = 2 (B) f’(1) = 2 and g’(1) = 1
(C) f’(1) = 1 and g’(1) = 0 (D) f’(1) = 1 and g’(1) = 1
Q.29 Water is flowing out at the rate of 6 m3/min from 1 1
a reservoir shaped like a hemispherical bowl of radius R
− (x2 − 1)y "+ xy '
Q.6 If y m + y m =2x, then the value of
= 13 m. The volume of water in the hemispherical bowl is equal to value equal to y
π 2
is given by V = . y (3R – y) when the water is y meter
deep . Find 3 (A) 4 m2 (B) 2 m2 (C) m2 (D) – m2
(a) At what rate is the water level changing when the  d 
water is 8 m deep? Q.7 If y2 = P(x), is a polynomial of degree 3, then 2  
 dx 
(b) At what rate is the radius of the water surface  3 d2 y 
 y . 2  equals:
changing when the water is 8 m deep?  dx 

(A) p”’(x) + P’(x) (B) P”(x).P”’(x)
Exercise 2 (C) P(x).P”’(x) (D) a constant

Methods of Differentiation − x3
Q.8 Given f(x) = + x2 sin 1.5 a – x sin a . sin 2a – 5
3
Single Correct Choice Type are sin (a2– 8a + 17) then :

x x x x x x dy (A) f(x) is not defined at x = sin 8

Q.1 If y = ….∞, then
a+ b+ a+ b+ a+ b+ dx (B) f ‘ (sin 8) > 0
a b (C) f ‘ (x) is not defined at x = sin 8
(A) (B)
ab + 2ay ab + 2by
(D) f ‘ (sin 8) < 0
a b
(C) (D)
ab + 2by ab + 2ay cos6x + 6 cos 4x + 15cos2x + 10 dy
Q.9 If y = , then =
cos5x + 5cos3x + 10 cos x dx

(A) 2 ain x + cos x (B) – 2 sin x

(C) cos 2x (D) sin 2x
2 1 . 4 4 | Methods of Differentiation and Applications of Derivatives

Q.10 A curve is parametrically represented by y = R(1 – Multiple Correct Choice Type

d2 y
cos θ) & x = R(θ – sinθ), then at θ = π is -
dx2  1 + x2 − 1 
 
Q.18 Let f(x) = tan–1  x  , then
1 1 1 1  
(A) – (B) (C) (D) –
2R 4R 2R 4R
(A) f(x) = 112 tan–1x ∆ x∈R – {0}
1
Q.11 If f(x) = (1 + x)n then the value of f(0) + f ‘ (0) + …. (B) f ‘ (x) = ζ x∈R – {0}
2(1 + x2 )
f n (0)
+ is – (C) f(x) is an odd function
n!
(A) n (B) 2n (C) 2n–1 (D) None (D) f(x) + f(– x) = p

dy
Q.19 If y = tan x tan 2x tan 3x then has the value to:
Q.12 If the function y = e + 2e is a solution of the
4x –x dx
d3 y dy (A) 3 sec2 3x tan x tan 2x + sec2 x tan 2x tan 3x + 2 sec2
− 13
3 dx
differential equation dx = K. then the value of K 2x tan 3x tan x
y
(B) 2y (cosec 2x + 2 cosec 4x + 3 cosec 6x)
(A) 4 (B) 6 (C) 9 (D) 12 (C) 3 sec2 3x – 2 sec22x – sec2 x
(D) sec2 x + 2 sec2 2x + 3 sec2 3x
d2 y
Q.13 x4 + 3x2y2 + 7xy3 + 4x3y – 15y4 = 0, then at
(1, 1) is - dx2 dy
Q.20 Let y = √x + Öx + x + ......∞ then
dx
(A) 2 (B) 1 (C) 7 (D) 0 1 x 1 y
(A) (B) (C) (D)
2y − 1 x − 2y 1 + 4x 2x + y
x
Q.14 If f(x) = ee . Let g(x) be it’s inverse then g’ (x) at
x = 2 is - dy
Q.21If 2x + 2y = 2x+y then has the value equal to
dx
n2 1 2y 1 2x (1 − 2y )
(A) (B) (C) 2n2 (D) e2 (A) – (B) (C) 1 − 2y (D)
2 2n2 x
2 1 − 2x 2y (2x − 1)
 1 − 2n | x |   3 + 2n | x | 
Q.15 y=tan–1  1 + 2n | x |  +tan–1  1 − 6n | x |  , then dy
    Q.22 If y + x + y – x = c, then is equal to
dx
d2 y
equals 2x x
dx2 (A)
2
(B)
c y + y 2 − x2
(A) 2 (B) 1 (C) 0 (D) –1
y − y 2 − x2 c2
xx
(C) (D)
Q.16 lim (x − x x ) equals - x 2y
x →0 +
(A) 1 (B) –1 (C) 0 (D) None of these
Application of Derivatives

Q.17 lim {(cot x)x + (1 – cos x)cosecx} is equal to - Single Correct Choice Type
x →0
(A) 2 (B) +1
3 2
(C) 0 (D) None of these Q.1 The line y = – x and y = – x intersect the curve
2 5
3x2 + 4xy + 5y2 – 4 = 0 at the  P(x1, y1) 2
), )
point P and O respectively. 5 2
Q (x2, y2)
The tangent drawn to the 3
curve at P and Q 2
3
(A) Intersect each other at angle 2
), )
of 45º 5
 M a them a ti cs | 21.45

(B) Are parallel to each other x

n
Q.8 The equation of the tangent to the curve   +
(C) Are perpendicular to each other n a
y
(D) None of these   = 2 (n ∈ N) at the point with abscissa equal to
b
‘a’ can be
Q.2 A curve is represented by the equations x = sec2 t
and y = cot t where t is a parameter. If the tangent at x y x y
(A)   +   = 2 (B)   –   = 2
the point P on the curve where t = π/4 meets the curve  a b  a b
again at the point Q then | PQ | is equal to
x y x y
5 3 5 5 (C)   –   = 0 (D)   +   = 0
(A) (B)  a b  a b
2 2
2 5 3 5
(C) (D) Multiple Correct Choice Type
3 2
 x35 x x 1
if x ≤ 1 Q.9 If + = 1 is a tangent to the curve x = Kt, y = ,
Q.3 Let f(x) =  then the number of a b t
3
−(x – 2) if x > 1 K > 0 then
critical points on the graph of the function is
(A) a > 0, b > 0 (B) a > 0, b < 0
(A) 1 (B) 2 (C) 3 (D) 4
(C) a < 0, b > 0 (D) a < 0, b < 0

Q.4 At any two points of the curve represented dy

parametrically by x = a(2 cos t –cos 2t), y = a(2 sin t – sin 2t) Q.10 The abscissa of the point on the curve = a + x,
dx
the tangents are parallel to the axis of x corresponding the tangent at which cuts off equal intersects from the
to the values of the parameter t differeing from each co-ordinate axes is (a > 0)
other by
a a
(A) 2π/3 (B) 3π/4 (C) π/2 (D) π/3 (A) (B) – (C) a 2 (D) – a 2
2 2

Q.5 At the point P (a, a”) on the graph of y = xn(n ∈ Q.11 The parabola y = x2 + px + q cuts the straight line y
N) in the first quadrant a normal is drawn. The normal = 2x – 3 at a point with adscissa 1. If the distance between
1 the vertex of the parabola and the x-axis is least then
intersects the y-axis at the point (0, b). If lim b = ,
then n equal
a→ 0 2
(A) p = 0 & q = – 2
(A) 1 (B) 3 (C) 2 (D) 4 (B) p = – 2 & q = 0
(C) least distance between the parabola and x–axis is 2
 −x2 for x > 0
Q.6 Let f (x) =  . Then x intercept of (D) least distance between the parabola and x–axis is 1
2
 x + 8 for x ≥ 0
the line that is tangent to the graph of f(x) is Q.12 The co-ordinates of the point(s) on the graph ……
(A) zero (B) – 1 (C) – 2 (D) – 4 x3 5x2
function, f(x) = – + 7x – 4 where the tangent
3 2
Q.7 The ordinate of all points on the curve drawn cut off intercepts from the co-ordinate axes
which are equal in magnitude but opposite in sign, is
1 (A) (2, 8/3) (B) (3, 7/2)
y= where the tangent is horizontal, is -
2sin x + 3cos2 x
2
(C) (1, 5/6) (D) None of these
(A) Always equal to 1/2
(B) Always equal to 1/3 Q.13 Equation of a tangent to the curve y cot x = y3 tan
π
(C) 1/2 or 1/3 according as n is an even or an odd x at the point where the abscissa is is
4
integer
(A) 4x + 2y = π + 2 (B) 4x – 2y = π + 2
(D) 1/2 or 1/3 according as n is an odd or an odd integer
(C) x = 0 (D) y = 0
2 1 . 4 6 | Methods of Differentiation and Applications of Derivatives

Q.14 The angle made by the tangent of the curve x = (v > u). At what point on the shore should be land so that
a(t + sin t cos t) ;y = a(1 + sin t)2 with the x-axis at any he reaches his house in the shortest possible time?(1983)
point on it is -
1 1 − sint 1 1 + sint Q.5 Find the coordinates of the point on the curve y =
(A) (π + 2t) (B) (C) (2t – π) (D)
4 cos t 4 cos2t x
, where the tangent to the curve has the greatest
1 + x2
Q.15 Consider the curve represented parametrically by the slope. (1997)
equation x = t3 – 4t2 – 3t and y = 2t2+ 3t – 5 where t ∈ R
If H denotes the number of point on the curve where Q.6 Find all the tangents to the curve y = cos (x + y), –
the tangent is horizontal and V the number of point 2π≤ x 2π, that are parallel to the line x + 2y = 0 (1997)
where the tangent is vertical then
(A) H = 2 and V = 1 (B) H = 1 and V = 2 Q.7 Find the point on the curve 4x2 + a2y2 = 4a2, 4 < a2<
(C) H = 2 and V = 2 (D) H = 1 and V = 1 8 that is farthest from the point (0, –2). (1987)

Q.8 Three normals are drawn from the point (c, 0) to the
Previous Years’ Questions 1
curve y2 = x. Show that c must be greater than . One
normal is always the x-axis. Find c for which the2 other
Q.1 If the line ax + bx + c = 0 is a normal to the curve two normals are perpendicular to each other  (1991)
xy = 1, then (1986)
(A) a > 0 , b > 0 (B) a > 0, b < 0 Q.9 What normal to the curve y = x2 forms the shortest
chord? (1992)
(C) a < 0, b > 0 (D) a < 0, b < 0

Q.10 Find the equation of the normal to the curve y =

Q.2 On the ellipse 4x2 + 9y2 = 1, the point at which the
(1 + x)y + sin–1(sin2x) at x = 0 (1993)
tangents are parallel to the line 8x = 9y, are  (1999)

2 1  2 1  2 1 2 1 Q.11 Tangent at a point P1{other than (0, 0)} on the

(A)  ,  (B)  − ,  (C)  − , −  (D)  , − 
5 5  5 5  5 5 5 5 curve y = x3 meets the curve again at P2. The tangent
at P2 meets the curve at P3, and so on. Show that the
abscissae of P1, P2, P3, …. < Pn, form a GP. Also find the
Q.3 Let C be the curve y3 – 3xy + 2 = 0. If H is the set of
ratio. [Area (dP1P2P3)] / [area (dP2P3P4)]  (1993)
points on the curve C where the tangent is horizontal
and V is the set of points on the curve C where the
tangent is vertical, then H = ………. and V = ……..[1997] Q.12 A rectangular sheet of fixed perimeter with sides
having their lengths in the ratio 8 : 15 is converted
an open rectangular box by folding after removing
Q.4 A swimmer S is in the sea at a distance d km from
square of equal area from all four corners. If the total
the closest point A on a straight shore. The house of the
area of removed squares is 100, the resulting box has
swimmer is on the shore at a distance L km from A. He can
maximum volume. Then the lengths of the sides of the
swim at a speed of u km/h and walk at a speed of v km/h
rectangular sheet are  (2013)
(A) 24 (B) 32 (C) 45 (D) 60

Q.13 Match List I with List II and select the correct answer using the code given below the lists : (2013)

List - I List – II
P 1/2 1.
 1  cos(tan−1 y) + y sin(tan−1 y)   1 5
   + y4  2 3
 2 −1 −1  
 y  cot(sin y) + tan(sin y)  

Q x−y 2.
If cos x + cos y + cos z = 0 = sin x + sin y + sin z then possible value of cos is 2
2
 M a them a ti cs | 21.47

R 3. 1
π  π 
If cos  − x  cos2 x + sin x sin2x
= sec x cos x sin 2x sec x + cos  + x  cos2x cos 2 x 2
4  4 
then possible value of sec x is

S 4. 1
If cot  sin−1 1 − x2  =
 
( )
sin tan−1 (x 6 ) , x ≠ 0 , then possible value of x is

(A) P → 4 ; Q → 3 ; R → 1 ; S → 2 (B) P → 4 ; Q → 3 ; R → 2 ; S → 1

(C) P → 3 ; Q → 4 ; R → 2 ; S → 1 (D) P → 3 ; Q → 4 ; R → 1 ; S → 2

PlancEssential Questions
JEE Main/Boards JEE Advanced/Boards
Exercise 1 Exercise 1

Methods of Differentiation Methods of Differentiation

Q.9 Q.12 Q.15 Q.23 Q.5 Q.8 Q.11 Q.12
Q.14 Q.18 Q.23
Application of Derivatives
Application of Derivatives
Q.7 Q.11 Q.15 Q.21
Q.8 Q.11 Q.15 Q.17
Q.20 Q.23 Q.24 Q.27
Exercise 2
Exercise 2
Methods of Differentiation
Q.2 Q.11 Q.12 Q.15 Methods of Differentiation

Q.17 Q.20 Q.1 Q.3 Q.5 Q.9

Q.11 Q.15 Q.20
Application of Derivatives
Q.5 Q.7 Q.8 Q.10 Application of Derivatives
Q.2 Q.4 Q.7 Q.11
Previous Years’ Questions
Q.2 Q.5 Q.7 Q.9 Previous Years’ Questions
Q.4 Q.9 Q.11
2 1 . 4 8 | Methods of Differentiation and Applications of Derivatives

Answer Key

JEE Main/Boards
Exercise 1
Methods of Differentiation

1 x +3 1
Q.1 e Q.2 cos (log x) Q.3 – sec3θ
2 x+3 x

−1 x
−1 mem tan 3x2 cos{log(x3 − 1)}
Q.4 Q.5 Q.6
2 x 1 − x 1 + x2 x3 − 1

Q.7 – e–x sin x Q.8 cot x . log2e Q.9 5ln sinx (cot x) (ln 5)

3 −1 sec2 x 1 − tanx
Q.10 − Q.11 Q.12
2 2
1 + x (1 − tanx) 2 1 + tanx

− cos cos x sin x 1  1 

Q.13 Q.24  
4 x cos x 2 x  1 + x2 
−1 x  −1 log(sinx)  2
Q.15 (sinx)cos cos x.cot x −  Q.16 1 Q.17
 1 + x2  1 + x2

dy 1 –2 1 1 1
Q.18 =– Q.19 Q.20 4 × ×
dx 2 1 − x2 1 − a2 x2 a+ a+ x a+ x

x cot  1 + x2 
1 1  
Q.21 Q.22 Q.23
1 + x2 a2 + x2 1+x 2

Application of Derivatives

1
Q.1 (2, 1) Q.2 90º Q.3 –
3
dy  3 −17 
Q.4 is not defined Q.5  ,  Q.6 (2 – 36)
dx 2 2 
Q.7 x + 12y – 21 = 0; 12x – y + 38 = 0 Q.8 2 bx + 2 ay – ab Q.9 M1M2= –1

Q.10 8x – 2y + 1 = 0 Q.11 4x – 4y + 33 = 0 Q.12 k2x + y – 2ck = 0

1 1 
Q.13 3 Q.14  ,  Q.15 (±3, 0)
2 4
Q.161/818.x + 9y – 55 = 0 ; x + 9y – 35 = 0 Q.17 -1 Q.18 55

x x
Q.19 20 x + y – 140 = 0 Q.20 + = a Q.21 x + 3y = 8; x + 3y = – 8
x1 y1

 −2 4  3a1/3b1/3
Q.22 2x + 2y = a 2
Q.23 (2, –4);  ,  Q.24 tan –1

 3 27  2(a2/3 + b2/3 )
 M a them a ti cs | 21.49

Exercise 2

Methods of Differentiation

Single Correct Choice Type

Q.1 B Q.2 B Q.3 D Q.4 D Q.5 A Q.6 B
Q.7 C Q.8 A Q.9 A Q.10 B Q.11 C Q.12 D
Q.13 D Q.14 D Q.15 B Q.16 B Q.17 C Q.18 A
Q.19 B Q.20 D Q.21 B

Application of Derivatives

Single Correct Choice Type

Q.1 B Q.2 D Q.3 C Q.4 D Q.5 D Q.6 A
Q.7 C Q.8 A Q.9 D Q.10 B Q.11 C Q.12 D
Q.13 D Q.14 D Q.15 C

Previous Years’ Questions

Q.1 C Q.2 A Q.3 D Q.4 D Q.5 D Q.6 A
Q.7 C Q.8 D Q.9 A Q.10 A Q.11 D

JEE Advanced/Boards
Exercise 1
Methods of Differentiation
4x3
Q.1 k =1, –1 or 0 Q.2 Q.3 9
−9 9
Q.4 Q.56 Q.6 25
4
 x cos x − sinx
if x ≠ 0 1
Q.7 k = 2 Q.9 f ‘ (x) =  x2 ; f “(0) = –
 3
 0 if x = 0
Q.11 Zero Q.13 3 Q.14 100

Q.15 6 Q.16 2(1 + 2x) . cos2(x + x2) Q.17 f(0) + kx

1 1
Q.18 – Q.19 = X[S1 t2 X 2 − S2 t1 X 2 ] + X3 S1 t1
2
1 + (x + n) 1 + x2 S2 t2
1 − 2x 1 1
Q.20 0 Q.21 Q.22 or –
2 1−x 2 2 2
Q.23 L.H.S = R.H.S Q.24 (a) (i) y’(0) = –1 ; (ii) y”(0) = 2; (b) a = 1, – 2

Application of Derivatives
−5
Q.1 2 3 x – y = 2 ( 3 − 1) or 2 3 x + y = 2 ( 3 + 1) Q.2 a = 1, b =
2
Q.3 (0, 1) Q.4 x = y – 1 = 0 Q.5 T : x – 2y = 0; N : 2x + y = 0
2 1 . 5 0 | Methods of Differentiation and Applications of Derivatives

82.73
Q.6 – Q.7 x + 2y = π/2 & x + 2y = –3π/2 Q.8 5
3
80
Q.9 (i) (a) 6.05, (b) ; (ii) 9.72 pcm2 Q10 a = 1 Q.11 1/16
27
1
Q.12– ; x – 4y = 2 Q.13 a = –1/2; b = – ¾; c = 3 Q.14 3, 12
x+2
−x
1
Q.15 9 Q.16 2e 2 Q.17 (a) n = –2, (c) ±
2 2
Q.18 65 Q.19 64 Q.20 (i) 1, (ii) 2

Q.21 (4, 11) & (– 4, – 31/3) Q.22 (i) 6 km/h; (ii) 2 km/hr Q.23 1 + 36 π cu. cm / sec

2 1
Q.24 cm/sec Q.25 1/48 π cm/s Q.26 cm/sec
4π 4
66
Q.27 Q.28 (a) r = (1 + t)1/4, (b) t = 80
7
1 5
Q.29 (a) – m/min, (b) – m/min
24π 288π

Exercise 2

Methods of Differentiation

Single Correct Choice Type

Q.1 D Q.2 B Q.3 C Q.4 B Q.5 D Q.6 C
Q.7 C Q.8 D Q.9 B Q.10 D Q.11 B Q.12 D
Q.13 C Q.14 C Q.15 C Q.16 B Q.17 B

Multiple Correct Choice Type

Q.18 B, C Q.19 A, B, C Q.20 ACD Q.21 A, B, C, D Q.22 A, B, C

Application of Derivatives

Single Correct Choice Type

Q.1 C Q.2 D Q.3 C Q.4 A Q.5 C Q.6 B
Q.7 D Q.8 A

Multiple Correct Choice Type

Q.9 A, D Q.10 A, B Q.11 B, D Q.12 A, B Q.13 A, B, D Q.14 A, B
Q.15 B, D

Previous Years’ Questions

ud
Q.1 B, C Q.2 B, D Q.3 H = ∅, V = {1.1} Q.4 Q.5 (0, 0)
v − u2
2

π 3π
Q.6 x + 2y = and x + 2y = – Q.7 (0, 2) Q.8 3/4 Q.9 2 x – 2y + 2 = 0
2 2
Q.10 dy/dx = 1 Q.11 1/16 Q.12 A C Q.13 B