SUBJECT NAME: ENGINEERING MATHEMATICS II

(COMMON TO BIO GROUPS)

SUBJECT CODE: SMT1106

COURSE MATERIAL

UNIT-II DIFFERENTIAL CALCULUS

____________________________________________________________________________

Definition 1. Differentiation
The rate at which a function changes with respect to the independent variable is called the
derivative of the function.
(i.e) If y= f(x) be a function, where x and y are real variables which are independent and
dy
dependent variables respectively, then the derivative of y with respect to x is dx.
Definition 2. Derivative of addition or subtraction of functions
d[f(x) ± g(x)] d[f(x)] d[g(x)]
If f(x) and g(x) are two functions of x, then = ±
dx dx dx

Definition 3. Product rule

d[uv] d[u] d[v]
If y = uv, where u and v are functions of x, then dx
=v dx
+ u dx

Definition 4. Quotient rule

du dv
u d u v −u
dx dx
If 𝑦 = , where u and v are functions of x,
v
then dx [v] = v2
Important Derivatives Formulae
1.
d
c   0 where ‘c’ is any constant.
dx

2.
d n
dx
 
x  nx n1 .

3.
d
log e x   1 .
dx x

4.
d x
dx
 
a  a x log a

5.
d x
dx
 
e  ex .

6.
d
sin x   cos x .
dx

7.
d
cos x    sin x .
dx

8.
d
tan x   sec 2 x .
dx
 9.
d
cot x    cos ec 2 x .
dx

10.
d
sec x   sec x tan x .
dx

11.
d
cos ecx    cos ecx cot x .
dx

12.
d

sin 1 x 
1
 .
dx 1 x2
1
13.
d

cos 1 x   .
dx 1 x2

14.
d
dx

tan 1 x 
1

1 x2
.

1
15.
d
dx

cot 1 x  
1 x2
.

16.
d

sec 1 x  
1
.
dx x 1 x2
1
17.
d

cos ec 1 x  
dx x 1 x2
Problems
I. Ordinary Differentiation Problems
𝟏
1. Differentiate 𝐱 + 𝐱
1
Solution Let 𝑦 = x + x
1
dy d(x + ) d(x) d(x−1 ) 1
x
Then dx = dx
= dx
+ dx
=1 − x2

2. Differentiate 𝟑𝐭𝐚𝐧 𝐱 + 𝟐 𝐜𝐨𝐬 𝐱 − 𝐞𝐱 + 𝟓

Solution:
Let y = 3tan x + 2 cos x − ex + 5
dy d(3tan x+2 cos x−ex +5) d(tan x) d(cos x) d(ex ) d(5)
Then = =3 +2 − +
dx dx dx dx dx dx
2 x
= 3𝑠𝑒𝑐 x − 2 sin x − e

3. Differentiate 𝐲 = 𝐞𝟐𝐱 𝐜𝐨𝐬𝟑𝐱

dy d(e2x cos3x) d(e2x ) d(cos3x)
Solution: dx
= dx
= cos3x dx + e2x dx
= 2cos3x e2x − 3e2x sin3x
4. Differentiate 𝐲 = 𝐞𝐬𝐢𝐧𝐱 𝐱 𝟐
dy d(esinx x2 )
Solution: dx = dx
 d(esinx ) d(x2 )
= x2 dx
+ esinx dx
= x 2 esinx (cosx) + 2xesinx
5. Differentiate y = 𝐱 𝟑 𝐞−𝐱 𝐭𝐚𝐧𝐱
dy d( x3 e−x tanx)
Solution: dx = dx
d( x3 ) d( e−x ) d(tanx)
= e−x tanx dx + x 3 tanx dx + x 3 e−x dx
2 −x 3 −x 3 −x 2
= 3x e tanx − x e tanx + x e sec x
𝐞𝐱
6. Differentiate 𝐲 = 𝐜𝐨𝐬𝐱
ex
dy d( ) cosx ex −ex (−sinx)
cosx
Solution: dx
= dx
= cos2 x
cosx ex +ex (sinx)
= cos2 x
𝐚𝐱+𝐛
7. Differentiate 𝐲 = 𝐜𝐱+𝐝
dy (cx+d)a−(ax+b)c
Solution:dx = (cx+d)2
(by quotient rule)
𝐱 𝟐 +𝟐𝐱+𝟑
8. Differentiate
√𝐱
1 −1 1
dy √x (2x+2)−(x2 +2x+3) x ⁄2 2√x (x+1)−(x2 +2x+3)
2 2√x
Solution: dx
= 2 = 2
(√x) (√x)
2√x ×2√x (x+1)−(x2 +2x+3) 4x(x+1)−(x2 +2x+3)
= 2 = 3
2√x (√x) 2x ⁄2
4x2 +4x−x2 −2x−3 3x2 +2x−3
= 3 = 3
2x ⁄2 2x ⁄2
9. Differentiate 𝐲 = (𝟑𝐱 𝟐 − 𝟏)𝟑
Solution: Given y = (3x 2 − 1)3
Differentiating w.r.to x, we get
dy
⇒ = 3(3x 2 − 1)2 6x
dx
= 3(9x 4 − 6x 2 + 1) = 27x 4 − 18x 2 + 3
𝟏+𝐬𝐢𝐧𝐱
10. Differentiate: 𝐥𝐨𝐠 (𝟏−𝐬𝐢𝐧𝐱)
1+sinx
Solution: Let y = log (1−sinx)
⇒ 𝑦 = log(1 + sinx) − log(1 − sinx)
Differentiate y w.r.to x, we get
dy 1 1
dx
= 1+sinx cosx − 1−sinx (−cosx)

(1−sinx)cosx+cosx(1+sinx)
= (1+sinx)(1−sinx)

𝑐𝑜𝑠𝑥−𝑠𝑖𝑛𝑥 𝑐𝑜𝑠𝑥+𝑐𝑜𝑠𝑥+ 𝑐𝑜𝑠𝑥 𝑠𝑖𝑛𝑥

= 1−𝑠𝑖𝑛2 𝑥
2 𝑐𝑜𝑠𝑥 1
= 𝑐𝑜𝑠2 𝑥
= 2 𝑐𝑜𝑠𝑥 = 2 𝑠𝑒𝑐𝑥

II. Differentiation Problems on Logarithmic Functions

1. Differentiate 𝐱 𝐬𝐢𝐧𝐱
Solution: Let y = x sinx
Taking log on both sides, we get logy = sinx logx
 Now differentiating with respect to x
1 dy 1
⇒ y dx
= logx(cosx) + sinx x (Using product rule)
dy 1
⇒ dx = y (logx( cosx) + sinx x)
dy y(xcosx logx+sinx)
⇒ dx = x
𝑑𝑦 sinx xcosx logx+sinx
⇒ 𝑑𝑥 = x ( x
)
𝐝𝐲 𝐥𝐨𝐠𝐱
2. If 𝐱 𝐲 = 𝐞𝐱−𝐲, prove that 𝐝𝐱 = (𝟏+𝐥𝐨𝐠𝐱)𝟐
y x−y
Solution: Given x = e
Taking log on both sides, we get logx y = logex−y
⇒ ylogx = (x − y)log e e
⇒ ylogx = (x − y)………(1)
1 dy dy
⇒ x y + logx dx = 1 − dx
dy dy y
⇒ logx + =1−
dx dx x
dy x−y
⇒ dx
(logx + 1) = x
dy x−y
⇒ dx = x(1+logx)
dy ylogx
⇒ dx = x(1+logx)….(2)
Again from (1) y + ylogx = x
y 1
⇒ y(1 + logx) = x, x = 1+logx
dy logx
⇒ =
dx (1+logx)2
…∞ 𝐝𝐲
𝐱𝐱
3. If y = 𝐱 , then find
𝐝𝐱
Solution:
x…∞
Given y = x x = xy
Taking log on both sides
logy = ylogx
Differentiating w. r. to x we get
1 dy 1 dy
⇒ y dx = y x + logx dx
1 dy y
⇒( − logx) =
y dx x
1−ylogx dy y
⇒( y
) dx =x
dy y y y2
⇒ = ( ) =
dx x 1−ylogx x(1−ylogx)

𝐱 𝟐 +𝟏
4. Differentiate 𝐲 = 𝐥𝐨𝐠 (𝐱 𝟐 −𝟏)
Solution:
y = log(x 2 + 1) − log(x 2 − 1)
dy 1 1
⇒ = 2x − 2 2x
dx x2 +1 x −1
dy 1 1
⇒dx = 2x (x2 +1 − x2 −1)
 dy x2 −1−(x2 +1) x2 −1−x2 −1) −2) −4x
⇒dx = 2x ((x2 +1)(x2 −1)) = 2x ( x4 −1
) = 2x (x4 −1) = x4 −1
𝟐
5. Differentiate 𝐲 = 𝐞𝟑𝐱 +𝟐𝐱+𝟑
dy 2
Solution: dx = e3x +2x+3 (6x + 2)

III. Differentiation of Implicit functions

If two variables x and y are connected by the relation f(x, y) = 0 and none of the variable is
directly expressed in terms of the other, then the relation is called an implicit function.

Problems
𝒅𝒚
1. Find 𝒅𝒙 , if 𝒙𝟑 +𝒚𝟑 = 𝟑𝒂𝒙𝒚
Solution:
Differentiating w.r.to x, we get
dy dy
⇒ 3x 2 + 3y 2 dx = 3a [x dx + y]
dy dy
⇒ 3y 2 dx − 3ax dx = 3ay − 3x 2
dy
⇒ (3y 2 − 3ax) = 3ay − 3x 2
dx
dy (3ay−3x2 ) 3(ay−x2 ) (ay−x2 )
⇒ = = =
dx 3y2 −3ax 3(y2 −ax) (y2 −ax)
𝐝𝐲
2. Find 𝐱 𝟐 + 𝐲 𝟐 = 𝟏𝟔
, 𝐢𝐟
𝐝𝐱
Solution:
Given x 2 + y 2 = 16
⇒ y 2 = 16 − x 2
⇒ y = √16 − x 2
dy 1 −1⁄
⇒ dx = 2 (16 − x 2 ) 2 × (−2x)
dy x x
⇒ dx = − = −y
√16−x2
𝒅𝒚
3. Find 𝒅𝒙, if 𝐱 = 𝐚𝐭 𝟐 , 𝐲 = 𝟐𝐚𝐭
Solution: Given x = at 2 , y = 2at
dx dy
dt
= 2at, dt = 2a
dy dy dx 2a 1
Now = / = =
dx dt dt 2at t
𝐝𝐲 𝟐 𝟑
4. Find , if 𝐲 + 𝐱 − 𝐱𝐲 + 𝐜𝐨𝐬𝐲 = 𝟎
𝐝𝐱
Solution:
Given y 2 + x 3 − xy + cosy = 0
dy d dy
⇒ 2y dx + 3x 2 − dx (xy) − siny dx = 0
dy dy
⇒ (2y − siny) dx + 3x 2 − (x dx + y × 1) = 0
dy
⇒ (2y − siny − x) dx + 3x 2 − y = 0
dy
⇒ (2y − siny − x) = y − 3x 2
dx
dy y−3x2
⇒ =
dx 2y−siny−x
 IV.Successive Differentiation
The process of differentiating a given function again and again is called as successive
differentiation and the results of such differentiation are called successive derivatives.
Notations:
dy d 2 y d 3 y th
dny
i) , , ,....... n order derivative: n
dx dx 2 dx 3 dx
ii) f ′(x) , f ′′(x) , f ′′′(x),…..., nth order derivative: f n
x 
iii) Dy,D2y,D3y……., nth order derivative: Dny
iv) y′ , y′′ , y′′′ ,……, nth order derivative: y(n)
v) y1,y2,y3,……..,nth order derivative: yn

Problems
d2y dy
1. If y=sin(sinx), prove that 2
+ tan x + y cos 2 x  0
dx dx
Solution:
Given y=sin(sinx)…………………(1)
Differentiating (1) with respect to x we get,
dy
= cos(sinx).cosx………………..(2)
dx
Differentiating (2) with respect to x we get,
d2y
=cos(sinx)(-sinx)+cosx(-sin(sinx).cosx) [Product Rule]
dx 2
d2y
= -sinxcos(sinx) – cos2xsin(sinx)……………..(3)
dx 2
Therefore,
d2y dy
2
+ tan x + y cos 2 x  -sinxcos(sinx) – cos2xsin(sinx) +(tanx) cos(sinx).cosx +y cos2x
dx dx
= - sinxcos(sinx) – y cos2x + sinxcos(sinx) + y cos2x

=0
d2y
2.If x = a(cost + tsint), y = a(sint – tcost), find
dx 2
Solution:
 a sin t  t cos t  sin t   at cos t.
dx
dt
 acos t  t sin t  cos t   at sin t.
dy
dt
dy
dy dt at sin t
   tan t.
dx dx at cos t
dt
d y d  dy  d  dy  dt
2
sec 2 t 1
       .
dx 2
dx  dx  dt  dx  dx at cos t at cos 3 t
 d2y h 2  ab
3. If ax  2hxy  by  1 then prove that 
2 2

dx 2 hx  by 3
Solution:
Given ax  2hxy  by  1 …………………………(1)
2 2

Differentiating (1) partially with respect to x,we get,

 dy  dy
2ax  2h x  y   2by  0
 dx  dx
dy  ax  hy 
Then ,  …………………………(2)
dx hx  by 
Differentiating (2) with respect to x again,we get,

hx  by  a  h dy   ax  hy h  b dy 

 dx   dx 
2
d y

dx 2 hx  by 2
h 2
 ab y   dy 2

h  ab x  h 2

 dy 
 ab  y  x 
 dx   dx 
hx  by 2 hx  by 2

h 2 
 ab  y  x
ax  hy  
   
hx  by   h 2  ab ax 2  by 2  2hxy


hx  by 2 hx  by 3
d2y h 2  ab
Thus,  (from(1))
dx 2 hx  by 3
nth derivative of some standard functions:

S.No Y=f(x) dny

yn  n
 Dn y
dx
1. e mx m n e mx
2. ax  b m mm  1m  2.........m  n  1a n ax  b 
mn

3. 1  1n n!a n
ax  b ax  b n1
4. log ax  b  1n1 n  1!a n
ax  b n
5. sin ax  b  n 
a n sin   ax  b 
 2 
6. cosax  b  n 
a n cos  ax  b 
 2 
 3
4. Find yn, where y 
( x  1)( 2 x  1)
Solution:
Resolving into partial fractions,
2 1
y 
2x 1 x  1
2 1 2 n n!  1 n!
n n
 yn  
2 x  1n1 x  1n1
5. Find the nth derivative of log(9x2-1).
Solution:
Let y  log( 9 x  1)  log 3x  13x  1
2

 log 3x  1  log 3x  1

dn dn
Then y n  log 3x  1  n log 3x  1
dx n dx

 yn 
 1n1 n  1!3n   1n1 n  1!3n
3x  1n 3x  1n
7 x 5
6. Find yn, where y  e
Solution:
7 x 5
Let y  e
d n 7 x 5

d n 5 7x
    
n
5 d
Then y n  n e  n e e e n
e7x
dx dx dx
 yn  e 7 e
5 n 7x

Leibnitz formula for the nth derivative of a product

If u and v are functions of x,then
D n uv   u n v  nc1u n1v1  nc2u n2 v2  ........  ncr u nr vr  .......  uvn
Problems
2
7. Find the nth differential coefficient of x log x
Solution:
Take u=logx , v = x2
d n1 d n2
dn 2
x 
log x 
dn

log x x 2
 nc log x  d 2
x  nc   log x  d2 2
 
x
dx n1 dx n2
1 2
dx n dx n dx dx 2
( since all the other terms are zero)


 1n1 n  1! x 2  n 1n2 n  2!2 x   nn  1 1n3 n  3!2
x n x n1 2 x 
n2
 2 1 n  3!
n2

x n2
1 1
8. If y = x2ex, show that y n  n(n  1) y 2  n(n  2) y1  (n  1)( n  2) y where yn stands for
2 2
dny
dx n

Solution:
Take u =ex ,v = x2
d n1 x d n2
yn  
dn 2 x
x e  
dn x 2
e x   
nc e   d 2
 
x  nc2 n 2 e x 
d2 2
x 
dx n1
1
dx n dx n dx dx dx 2
( since all the other terms are zero)
 y n  e x x 2  2nxe x  nn  1e x

Now,
y1  x 2 e x  2 xex , y2  x 2 e x  4 xex  2e x

 nn  1 y 2  nn  2 y1  n  1n  2 y 

1 1 
nn  1 x 2 e x  4 xex  2e x 
2 2 2


 nn  2 x 2 e x  2 xex   n  1n  2x 2 e x
2
 nn  1 n  1n  2  xex 2nn  1  2nn  2  nn  1e x
 x 2e x   nn  2  
 2 2
=yn on simplification.
V. Partial Differentiation
Consider z = f(x, y), here z is a function of two independent variables x and y. z can be
differentiated with respect to x or y but when we are differentiating z with respect to x (or y ) we
must keep the variable y (or x ) as a constant.

Notations:
Let z=f(x,y)
First order partial derivatives of f(x, y) with respect to x and y.
∂f ∂f
= fx , = fy
∂x ∂y
Second order partial derivatives of f(x, y) with respect to x and y
∂2 f ∂2 f
= fxx , = fyy
∂x2 ∂y2
Second order mixed partial derivatives of f(x, y)
∂2 f ∂2 f
= fxy, = fyx
∂x ∂y ∂y ∂x
Problems:
𝛛𝐮 𝛛𝐮
1. If u = 𝐱 𝟑 + 𝐲 𝟑 + 𝟑𝐱𝐲, find 𝛛𝐱
, 𝛛𝐲
Solution: Given If u = x 3 + y + 3xy 3
∂u ∂u
∂x
= 3x 2 + 3y , ∂y = 3y 2 + 3x
𝛛𝐮 𝛛𝐮 𝛛𝐮 𝟑
2. If u = log (𝐱 𝟑 + 𝐲 𝟑 + 𝐳 𝟑 − 𝟑𝐱𝐲𝐳), show that 𝛛𝐱
+ 𝛛𝐲 + 𝛛𝐳
= (𝐱 +𝐲 +𝐳)
 Solution: u = log (x 3 + y 3 + z 3 − 3xyz)
∂u 1
∂x
= x3 + y3 + z3 −3xyz 3x 2 − 3yz ,
∂u 1
∂y
= x3 + y3 + z3 −3xyz 3y 2 − 3xz,
∂u 1
∂z
= x3 + y3 + z3 −3xyz 3z 2 − 3xy
∂u ∂u ∂u 3x2 +3y2 +3z2 −3yz−3xz−3xy
Now ∂x + ∂y + ∂z = x3 + y3 + z3 −3xyz
3(x2 + y2 + z2 −xy−yz−zx ) 3
= (x+y+z)(x2 + y2 + z2 −xy−yz−zx ) = x+y+z
𝟐 𝟐
3. 𝐈𝐟 𝐟(𝐱, 𝐲) = 𝐱 𝐬𝐢𝐧 𝐲 + 𝐲 𝐜𝐨𝐬 𝐱, then find its all first and 2nd order partial derivatives.
Solution: Given f(x, y) = x 2 sin y + y 2 cos x
fx = 2x sin y − y 2 sin x; fy = x 2 cos y + 2y cos x.
fxx = 2 sin y − y 2 cos x; fyy = −x 2 sin y + 2 cos x;
fxy = 2x cos y − 2y sin x; fyx = 2x cos y − 2y sin x.
𝐲
4. If 𝐟(𝐱, 𝐲) = 𝐱 𝐥𝐨𝐠 𝐱 , 𝐭𝐡𝐞𝐧 𝐟𝐢𝐧𝐝 𝐢𝐭𝐬 𝐚𝐥𝐥 𝟏𝐬𝐭 𝐚𝐧𝐝 𝟐𝐧𝐝 𝐨𝐫𝐝𝐞𝐫 𝐝𝐞𝐫𝐢𝐯𝐚𝐭𝐢𝐯𝐞𝐬.
y 1 –y y logx
Solution: fx = + logx ( )= (1 − logx), fy = ,
x x x2 x2 x
y 1 2y y y
fxx = (− x ) − (1 − log x) = x3 (−1 − 2 (1 − log x )) = (log x −3);
x2 x3 x3
1 1 1 1 1
fyy = 0, fyx = (1 − log x); fxy = x x – x2 log x = x2 (1 − log x).
x2
𝛛𝐮 𝛛𝐮 𝛛𝐮
5. Find , ,
𝛛𝐱 𝛛𝐲 𝛛𝐳
for 𝐮 = 𝐬𝐢𝐧(𝐚𝐱 + 𝐛𝐲 + 𝐜𝐳)
Solution:
∂u
∂x
= a cos(ax + by + cz)
∂u
= b cos(ax + by + cz)
∂x
∂u
∂x
= c cos(ax + by + cz)

VI. Euler’s Theorem for Homogeneous Functions

A homogenous function of degree n of the variables x, y, z is a function in which each term has
degree n. For example, the function f (x, y, z) = Ax3 + By3 + Cz3 + Dxy2 + Exz2 + Fyz2 + Gyx2 +
Hzx2 + Izy2 + Jxyz, is a homogeneous function of x, y, z, in which all terms are of degree three.
Note:
A function f(x ,y) of two independent variables x and y is said to be homogeneous in x and y of
degree n if f(tx, ty) = t n f(x, y)for any positive quantity t.

Euler’s theorem:
1). If f(x, y) is a homogeneous function of degree n, then
∂f ∂f
x ∂x + y ∂y = nf

2). If f(x, y, z) is a homogeneous function of degree n, then

∂f ∂f ∂f
x ∂x + y ∂y + z ∂z = nf

Result: If z is a homogeneous function of x, y of degree n and z=f(u) then

∂u ∂u f(u)
(i). x ∂x + y ∂y = n f′(u)

Problems on Euler’s theorem

1. Verify Euler’s theorem when 𝐮 = 𝐱 𝟑 + 𝐲 𝟑 + 𝐳 𝟑 + 𝟑𝐱𝐲𝐳
Solution:
Given u = x 3 + y 3 + z 3 + 3xyz
Now tu = (tx)3 + (ty)3 + (tz)3 + 3txtytz
= t 3 (x 3 + y 3 + z 3 + 3xyz) = t 3 u
Therefore u is a homogeneous function of degree 3.
∂u
∂x
= 3x 2 + 3yz
∂u
∂y
= 3y 2 + 3xz
∂u
∂z
= 3z 2 + 3xy
∂f ∂f ∂f
Therefore x +y + z = x(3x 2 + 3yz) + y(3y 2 + 3xz) + z(3z 2 + 3xy)
∂x ∂y ∂z
= 3x 3 + 3y 3 + 3z 3 + 9xyz
= 3(x 3 + x 3 + 3xy) = 3u
Hence Euler’s theorem is verified.
𝐲 𝛛𝐮 𝛛𝐮
2. If 𝐮 = 𝐱𝒍𝒐𝒈 (𝐱), then prove that 𝐱 𝛛𝐱 + 𝐲 𝛛𝐲 = 𝐧𝐮
Solution:
y
Given u = xlog (x)
u is a homogeneous function of degree 1.
∂u ∂u
Therefore by Euler’s theorem x ∂x + y ∂y = 1 × u = u
𝟏 𝟏 𝐥𝐨𝐠𝐱−𝐥𝐨𝐠𝐲 𝛛𝐟 𝛛𝐟
3. If (𝐱, 𝐲) = 𝐱 𝟐 + 𝐱𝐲 + 𝐱 𝟐 +𝐲 𝟐
, then prove that 𝐱 𝛛𝐱 + 𝐲 𝛛𝐲 + 𝟐𝐟 = 𝟎

Solution:
1 1 logx−logy
f(x, y) = 2 + + 2 2
x xy x +y
1 1 logtx−logty
Now f(tx, ty) = (tx)2 + txty + (tx)2 +(ty)2
tx
1 1 log
= + + 2 2 ty 2
t2 x2 2
t xy t (x +y )

1 1 1 logx−logy
= t2 (x2 + xy + x2 +y2
)

1 1 logx−logy
= t −2 ( 2 + + 2 2 )
x xy x +y
Therefore f(x, y) is a homogeneous function of degree -2
𝜕𝑓 𝜕𝑓
By Euler’s theorem , 𝑥 𝜕𝑥 + y 𝜕𝑦 = −2𝑓
𝜕𝑓 𝜕𝑓
⇒𝑥 𝜕𝑥 + y 𝜕𝑦 + 2𝑓 = 0
𝐱 𝟑 +𝐲 𝟑 𝛛𝐮 𝛛𝐮
4. If u = t𝐚𝐧−𝟏 ( 𝐱−𝐲
), show that 𝐱 𝛛𝐱
+ 𝐲 𝛛𝐲
= 𝐬𝐢𝐧𝟐𝐮
x3 +y3
Solution: Given u = tan−1 ( x−y
)
x3 +y3
⇒ tanu = ( x−y
)
x3 +y3
Let z = tanu = ( x−y )
And z is a homogeneous function of order 2.
 ∂u ∂u f(u)
We know that x ∂x + y ∂y = n f′(u)
Here f(u) = tanu
⇒ f ′ (u) = sec 2 u
Therefore by the result,
∂u ∂u tanu sinu
x ∂x + y ∂y = 2 sec2 u = 2 cosu × cos2 u
= 2sinu × cosu = sin2u
(Or)
∂z ∂z
By Euler’s theorem, x + y = nz
∂x ∂y
2 ∂u 2 ∂u
⇒ xsec u + ysec u = 2z
∂x ∂y
∂u ∂u
⇒ xsec 2 u + ysec 2 u = 2tanu
∂x ∂y
1 ∂u 1 ∂u sinu
⇒ x cos2 u ∂x + y cos2 u ∂y = 2 cosu
1 ∂u 1 ∂u sinu
⇒ x cosu ∂x + y cosu ∂y = 2 1
∂u ∂u
⇒ x ∂x + y ∂y = 2sinu cosu = sin 2u.

All the best