Di!

erentiation Practice Sheet Solutions

Vaibhav Dixit

Section A: Di!erentiation of Elementary Functions

Di!erentiate with respect to x.
1. y = x2 + x + 8
d d
Using the power rule, dx
(xn ) = nxn→1 , and the constant rule, dx
(c) = 0:
dy
= 2x + 1
dx
2. y = tan x + cot x
d d
Using dx (tan x) = sec2 x and dx
(cot x) = → csc2 x:
dy
= sec2 x → csc2 x
dx
3. y = sin x + cos x
d d
Using dx (sin x) = cos x and dx
(cos x) = → sin x:
dy
= cos x → sin x
dx

4. y = ln x + ex
d 1 d
Using dx (ln x) = x
and dx
(ex ) = ex :
dy 1
= + ex
dx x

5. y = x9
Using the power rule:
dy
= 9x8
dx
6. y = 3x10
Using the power rule and constant multiple rule:
dy
= 3 · 10x9 = 30x9
dx

7. y = 4x10
Using the power rule and constant multiple rule:
dy
= 4 · 10x9 = 40x9
dx
1
 8. y = 4x3/2
Using the power rule:
dy 3
= 4 · x1/2 = 6x1/2
dx 2

9. y = 12 x5/2
Using the power rule:
dy 1 5 5
= · x3/2 = x3/2
dx 2 2 4

10. y = x3 + x6
Using the power rule:
dy
= 3x2 + 6x5
dx

11. y = 3x2 + 2x3

Using the power rule:
dy
= 3 · 2x + 2 · 3x2 = 6x + 6x2
dx

12. y = 4x3 + 5x2 + 6

Using the power rule and constant rule:
dy
= 4 · 3x2 + 5 · 2x = 12x2 + 10x
dx

13. y = ax2 + bx + c (where a, b, c are constants)

Using the power rule and constant rule:
dy
= 2ax + b
dx

14. y = 3x4/3 + 4x3/4 + 5x

Using the power rule:
dy 4 3
= 3 · x1/3 + 4 · x→1/4 + 5 = 4x1/3 + 3x→1/4 + 5
dx 3 4

15. y = x + x1
Rewrite as y = x + x→1 . Using the power rule:
dy 1
= 1 + (→1)x→2 = 1 → 2
dx x

16. y = x2 + x12
Rewrite as y = x2 + x→2 . Using the power rule:
dy 2
= 2x + (→2)x→3 = 2x → 3
dx x

2
17. y = 3x + x42 + 5x
Rewrite as y = 3x + 4x→2 + 5x. Using the power rule:
dy 8
= 3 + 4 · (→2)x→3 + 5 = 8 → 3
dx x

18. y = (x → 1)(x2 + 1)
du dv
Using the product rule, let u = x → 1, v = x2 + 1. Then dx
= 1, dx
= 2x. Thus:

dy dv du
=u +v = (x → 1) · 2x + (x2 + 1) · 1 = 2x2 → 2x + x2 + 1 = 3x2 → 2x + 1
dx dx dx

3
19. y = xx3 →1
+1
du dv
Using the quotient rule, let u = x3 → 1, v = x3 + 1. Then dx
= 3x2 , dx
= 3x2 . Thus:
du dv
dy dx
v → u dx 3x2 (x3 + 1) → (x3 → 1)3x2 3x5 + 3x2 → 3x5 + 3x2 6x2
= = = =
dx v2 (x3 + 1)2 (x3 + 1)2 (x3 + 1)2

2
20. y = xx2 +2x+1
→2x+3
du
Using the quotient rule, let u = x2 + 2x + 1, v = x2 → 2x + 3. Then dx
= 2x + 2,
dv
dx
= 2x → 2. Thus:

dy (2x + 2)(x2 → 2x + 3) → (x2 + 2x + 1)(2x → 2)

=
dx (x2 → 2x + 3)2

Simplify the numerator:

(2x + 2)(x2 → 2x + 3) = 2x3 → 4x2 + 6x + 2x2 → 4x + 6 = 2x3 → 2x2 + 2x + 6

(x2 + 2x + 1)(2x → 2) = 2x3 + 4x2 + 2x → 2x2 → 4x → 2 = 2x3 + 2x2 → 2x → 2

Numerator = (2x3 →2x2 +2x+6)→(2x3 +2x2 →2x→2) = →4x2 +4x+8 = 4(→x2 +x+2)
Thus:
dy 4(→x2 + x + 2)
= 2
dx (x → 2x + 3)2
↑ ! "
21. y = ( x + 1) ↑1 →1
x
Rewrite as y = (x1/2 + 1)(x→1/2 → 1). Using the product rule, let u = x1/2 + 1,
v = x→1/2 → 1. Then du
dx
= 12 x→1/2 , dx
dv
= → 12 x→3/2 . Thus:
# $ # $
dy dv du 1/2 1 →3/2 →1/2 1 →1/2
=u +v = (x + 1) → x + (x → 1) x
dx dx dx 2 2
1 1 1 1 1 1
= → x→1 → x→3/2 + x→1 → x→1/2 = → x→3/2 → x→1/2
2 2 2 2 2 2
# $
1 1 1 1
= → 3/2 → 1/2 = → ↑ +1
2x 2x 2 x x

3
Section B: Di!erentiation by Product Rule
22. y = xex
du dv
Let u = x, v = ex . Then dx
= 1, dx
= ex . Using the product rule:
dy dv du
=u +v = xex + ex · 1 = ex (x + 1)
dx dx dx

23. y = sin x cos x

du dv
Let u = sin x, v = cos x. Then dx
= cos x, dx
= → sin x. Using the product rule:
dy
= sin x(→ sin x) + cos x(cos x) = cos2 x → sin2 x
dx
Alternatively, using the identity cos2 x → sin2 x = cos 2x:
dy
= cos 2x
dx

Section C: Di!erentiation by Quotient Rule

24. y = 2x+5
3x→2
du dv
Let u = 2x + 5, v = 3x → 2. Then dx
= 2, dx
= 3. Using the quotient rule:
du dv
dy dx
v → u dx 2(3x → 2) → (2x + 5) · 3 6x → 4 → (6x + 15) →19
= 2
= 2
= 2
=
dx v (3x → 2) (3x → 2) (3x → 2)2

25. y = lnxx
du
Let u = ln x, v = x. Then dx
= x1 , dv
dx
= 1. Using the quotient rule:
1
dy x
· x → ln x · 1 1 → ln x
= =
dx x2 x2

Section D: Di!erentiation by Chain Rule

26. y = sin 5x
du
Let u = 5x, so y = sin u, dx
= 5. Using the chain rule:
dy dy du
= · = cos u · 5 = 5 cos 5x
dx du dx

27. y = 2 sin(ωx + ε)
du
Let u = ωx + ε, so y = 2 sin u, dx
= ω. Using the chain rule:
dy
= 2 cos u · ω = 2ω cos(ωx + ε)
dx

28. y = (4 → 3x)9
du
Let u = 4 → 3x, so y = u9 , dx
= →3. Using the chain rule:
dy
= 9u8 · (→3) = →27(4 → 3x)8
dx
4
Section E: Maxima & Minima
29. A particle’s position is x = →t2 + 4t + 4. Find the maximum position.
The position function is a quadratic, x(t) = →t2 + 4t + 4. Since the coe”cient of t2
is negative, the parabola opens downward, and the maximum occurs at the vertex.
b
For f (t) = at2 + bt + c, the vertex is at t = → 2a . Here, a = →1, b = 4:

4
t=→ =2
2(→1)

Evaluate x at t = 2:

x(2) = →(2)2 + 4 · 2 + 4 = →4 + 8 + 4 = 8

Alternatively, find the derivative:

dx
= →2t + 4
dt
dx
Set dt
= 0:
→2t + 4 = 0 =↓ t = 2
d2 x
Second derivative: dt2
= →2 < 0, confirming a maximum. Thus, the maximum
position is:
x=8

30. Find the values of f (x) = 2x3 → 15x2 + 36x + 11 at maxima and minima.
Find critical points by computing the derivative:

f ↓ (x) = 6x2 → 30x + 36

Set f ↓ (x) = 0:

6x2 → 30x + 36 = 0 =↓ x2 → 5x + 6 = 0 =↓ (x → 2)(x → 3) = 0 =↓ x = 2, 3

Second derivative: f ↓↓ (x) = 12x → 30. Evaluate at critical points:

f ↓↓ (2) = 12 · 2 → 30 = →6 < 0 (maximum at x = 2)

f ↓↓ (3) = 12 · 3 → 30 = 6 > 0 (minimum at x = 3)

Evaluate f (x) at critical points:

f (2) = 2(23 ) → 15(22 ) + 36(2) + 11 = 16 → 60 + 72 + 11 = 39

f (3) = 2(33 ) → 15(32 ) + 36(3) + 11 = 54 → 135 + 108 + 11 = 38

Thus, the function values are:

Maximum at x = 2 : f (2) = 39, Minimum at x = 3 : f (3) = 38

5
Section F: Miscellaneous
31. Volume of a cylinder: V = ϑr2 h.
(a) Height increasing at dh
dt
= 5 m/s, radius constant ( dr
dt
= 0). Di!erentiate V :
dV dh
= ϑr2
dt dt
dV
= ϑr2 · 5 = 5ϑr2 m3 /s
dt
dr
(b) Radius increasing at dt
= 5 m/s, height constant ( dh
dt
= 0). Di!erentiate V :
dV dr dh
= ϑ(2r h + r2 ) = ϑ · 2r · 5 · h = 10ϑrh m3 /s
dt dt dt
dh dr
(c) Both increasing at dt = 5 m/s, dt = 5 m/s. Using the product rule:
dV dr dh
= ϑ(2r h + r2 ) = ϑ(2r · 5 · h + r2 · 5) = ϑ(10rh + 5r2 ) = 5ϑr(2h + r) m3 /s
dt dt dt

32. Ladder 5 m long, wall 4 m high, base pulled at 2 m/s. Find dω dt

when ϖ = 45↔ .
Let x be the distance from the wall to the ladder’s base, ϖ the angle between the
ladder and ground. The ladder’s length is 5 m, so by Pythagoras: x2 + y 2 = 25,
where y is the height. Also, sin ϖ = y5 , so y = 5 sin ϖ. Given dx
dt
= 2 m/s, find dω
dt
at
↔
ϖ = 45 .
Di!erentiate y = 5 sin ϖ:
dy dϖ
= 5 cos ϖ
dt dt
From x2 + y 2 = 25, di!erentiate:
dx dy dx dy
2x + 2y = 0 =↓ x + y =0
dt dt dt dt
dy
Substitute dt
:
dϖ dϖ 2x
x · 2 + y · 5 cos ϖ = 0 =↓ =→
dt dt 5y cos ϖ
↑ ↑ ↑
At ϖ = 45↔ , sin ϖ = cos ϖ = 22 , so y = 5 · 22 = 5 2 2 . Then:
& ) ↑ *2 +
' + ↑
% ' 5 2 25 · 2 50 5 2
x = 25 → y 2 = 25 →( = 25 → = =
2 4 4 2
↑ ↑ ↑ ↑ ↑
dϖ 2 · 522 5 2 5 2 5 2 2
=→ ↑ ↑ = → ↑ ↑ = →
25·2 = → =→ rad/s
dt 5· 5 2
· 2 25· 2· 2
4
12.5 5
2 2 4

33. Two cars: one west at 6 mph, one north at 3 mph. Distance after 3 hours?
Let the westbound
% car’s position↑ be (→6t,↑0), northbound car’s position be (0, 3t).
Distance s = (6t) + (3t) = 45t2 = 3 5t. Di!erentiate:
2 2

ds ↑
= 3 5 mph
dt
After 3 hours, the rate is still:
ds ↑
= 3 5 ↔ 6.708 mph
dt
6
34. Area A = ε4 D2 , find dD
dA
when D = 10 m.
Di!erentiate with respect to D:
dA ϑ ϑD
= · 2D =
dD 4 2
At D = 10:
dA ϑ · 10
= = 5ϑ m2 /m
dD 2