R S Aggarwal Solutions for Class 11 Maths Chapter 28 Differentiation Exercise 28E

Exercise 28E of Chapter 28 is based on topic “Derivative of a function of a function” or “Chain Rule”, which states that, if y = f(t) and t = g(x) then dy/dx = (dy/dt x dt/dx). This rule can also be extended as: If y = f(t), t = g(u) and u = h(x) then

Chain Rule

Download chapter 28 exercise 28E solutions from the link given below.

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rs aggarwal solution class 11 maths chapter 28 ex e
rs aggarwal solution class 11 maths chapter 28 ex e
rs aggarwal solution class 11 maths chapter 28 ex e
rs aggarwal solution class 11 maths chapter 28 ex e
rs aggarwal solution class 11 maths chapter 28 ex e

 

Access Answers to Maths R S Aggarwal Class 11 Chapter 28 Differentiation Exercise 28E Page number 892

Differentiate the following with respect to x:

Differentiate below functions using “Chain Rule”:

If y = f(t) and t = g(x) then dy/dx = dy/dt x dt/dx

Question 1: Differentiate sin 4x

Solution:

Let y = sin 4x

If 4x = t then sin4x = sin t or y = sin t

dy/dt = cos t

dt/dx = 4

Now,

dy/dx = dy/dt x dt/dx = 4 cos t

Substituting the value of t back, we get

dy/dx = 4 cos 4x

Question 2: Differentiate cos 5x

Solution:

Let y = cos 5x

If t = 5x then cos 5x = cos t or y = cos t

dy/dt = -sin t

dt/dx = 5

Now,

dy/dx = dy/dt x dt/dx = -5 sin t

Substituting the value of t back, we get

dy/dx = -5 sin 5x

Question 3: Differentiate tan 3x

Solution:

Let y = tan 3x

If t = 3x then tan 3x = tan t or y = tan t

dy/dt = sec2 t

dt/dx = 3

Now,

dy/dx = dy/dt x dt/dx = 3 sec2 t

Substituting the value of t back, we get

dy/dx = 3sec2 3x

Question 4: Differentiate cos x3

Solution:

Let y = cos x3

If t = x3 then cos x3= cos t or y = cos t

dy/dt = -sin t

dt/dx = 3x2

Now,

dy/dx = dy/dt x dt/dx = 3 x2 (-sin t)

Substituting the value of t back, we get

dy/dx = 3 x2 (-sin x3)

= -3 x2 sin x3

Question 5: Differentiate cot2 x

Solution:

Let y = cot2 x

dy/dx = – 2cotx [d/dx (cot x)]

= – 2 cot x cosec2x

Question 6: Differentiate tan3 x

Solution:

Let y = tan3 x

dy/dx = 3 tan2 x [d/dx (tan x)]

= 3 tan2 x sec2x

Question 7: Differentiate tan √x

Solution:

Let y = tan √x

dy/dx = sec2√x d/dx (√x)

= sec2√x {1/(2√x) }

= sec2√x / 2√x

Question 8: Differentiate ex^4

Solution:

R S Aggarwal Class 11 chapter 28 Ex 28E solution 8

Question 9: Differentiate ecotx

Solution:

R S Aggarwal Class 11 chapter 28 Ex 28E solution 9

Question 10: Differentiate √(sin x)

Solution:

d/dx √(sin x) = d/dx (sin x)1/2

= 1/2 * (sinx)-1/2 * d/dx (sin x)

= 1/2 * (sinx)-1/2 * cos x

= cosx / 2√(sin x)

Question 11: Differentiate (5 + 7x)6

Solution:

d/dx(5 + 7x)6 = 6(5 + 7x)5 * d/dx(5 + 7x)

= 6(5 + 7x)5 * 7

= 42(5 + 7x)5

Question 12: Differentiate (3 – 4x)5

Solution:

d/dx(3 – 4x)5 = 5(3 – 4x)4 * d/dx (3 – 4x)

= 5(3 – 4x)4 * ( – 4)

= – 20(3 – 4x)4

Question 13: Differentiate (3x2 – x + 1)4

Solution:

d/dx (3x2 – x + 1)4 = 4(3x2 – x + 1)3 * d/dx (3x2 – x + 1)

= 4(3x2 – x + 1)3 * (6x – 1)

= 4(3x2 – x + 1)3 (6x – 1)

Question 14: Differentiate (ax2 + bx + c)n

Solution:

d/dx (ax2 + bx + c)n = n(ax2 + bx + c)n-1 * d/dx (ax2 + bx + c)

= n(ax2 + bx + c)n-1 (2ax + b)

Question 15: Differentiate

R S Aggarwal Class 11 chapter 28 Ex 28E solution 15

Solution:

d/dx (x2 – x + 3)-3 = -3(x2 – x + 3)-4 * d/dx (x2 – x + 3)

= -3(x2 – x + 3)-4(2x2 – 1)

Question 16: Differentiate sin2 (2x + 3)

Solution:

d/dx sin2 (2x + 3) = 2 sin (2x + 3) * d/dx {sin(2x + 3)}

= 2 sin (2x + 3) * cos(2x + 3) * d/dx (2x + 3)

= 2 sin (2x + 3) * cos(2x + 3) * 2

= 4 sin(2x + 3) cos(2x + 3)

= 2 sin (4x + 6)

[Using identity: 2sin x cos x = sin 2x]

Question 17: Differentiate cos2(x3)

Solution:

d/dx cos2(x3) = 2 cos(x3) * d/dx { cos (x3)}

= 2 cos(x3) * ( – sin (x3)) * d/dx (x3)

= 2 cos(x3) * ( – sin (x3)) * 3x2

= – 6x2 cos(x3) sin x3

Question 18: Differentiate √(sin (x3 ))

Solution:

R S Aggarwal Class 11 chapter 28 Ex 28E solution 18

Question 19: Differentiate √(x sin x)

Solution:

R S Aggarwal Class 11 chapter 28 Ex 28E solution 19

Question 20: Differentiate

R S Aggarwal Class 11 chapter 28 Ex 28E question 20

Solution:

R S Aggarwal Class 11 chapter 28 Ex 28E solution 20


Access other exercise solutions of Class 11 Maths Chapter 28 Differentiation

Exercise 28A Solutions

Exercise 28B Solutions

Exercise 28C Solutions

Exercise 28D Solutions

R S Aggarwal Solutions for Class 11 Maths Chapter 28 Exercise 28E

Class 11 Maths Chapter 28 Differentiation Exercise 28E is based on the topic: Derivative of a function of a function. The step by step solution in R S Aggarwal Solution for Class 11 Mathematics offers invaluable help while doing homework and preparing for exams.

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