Chain Rule Questions

Chain rule questions and solutions are available here to help students learn how to find the derivative of a composition of functions using a simple technique. In this article, you will learn how to perform the differentiation of composite functions with the help of solved questions. You can also practise additional questions provided at the end of the page to test your understanding.

What is the Chain rule of differentiation?

Chain rule is one of the most important methods of finding the derivative of a composite function. Consider a function y = f(x), where x is another function say x = g(u), such that the derivative of y with respect to x is given by:

dy/dx = (dy/du) (du/dx)

This is the general formula used for the chain rule of differentiation. However, to find the derivative of a function using the chain rule, one must be aware of the basic differentiation formulas.

Learn more about the chain rule of differentiation here.

Chain Rule Questions and Answers

1. If y = cos x³, find dy/dx.

Solution:

Given,

y = cos x³

Let y = cos u and u = x³

dy/du = -sin u

du/dx = 3x²

Now, using the chain rule, we have;

dy/dx = (dy/du) (du/dx)

= (-sin u) (3x²)

= – 3x² sin x³ {since u = x³)

Therefore, dy/dx = -3x² sin x³.

2. Find the derivative of y = e^{x sin x}.

Solution:

Given function is:

y = e^{x sin x}

Let y = e^t and t = x sin x

dy/dt = e^t

dt/dx = (d/dx)(x sin x)

= x cos x + sin x {from the product rule of differentiation}

Now, using the chain rule, we have;

dy/dx = (dy/dt) (dt/dx)

= e^t (x cos x + sin x)

= e^{x sin x} (x cos x + sin x)

Thus, the derivative of y = e^{x sin x} is dy/dx = e^{x sin x} (x cos x + sin x).

3. What is the derivative of the function y = ln(x + x⁷)?

Solution:

Given function is:

y = ln(x + x⁷)

Let u = x + x⁷

And y = ln(u)

Now, du/dx = (d/dx)(x + x⁷) = 1 + 7x⁶

dy/du = (d/du) ln(u) = 1/u

Using the chain rule, we have;

dy/dx = (dy/du) (du/dx)

= (1/u) (1 + 7x⁶)

= (1 + 7x⁶)/(x + x⁷) {since u = x + x⁷}

Hence, the derivative of y = ln(x + x⁷) is dy/dx = (1 + 7x⁶)/(x + x⁷).

4. Find dy/dx if y = 2 sin(3 cos 4x).

Solution:

Given,

y = 2 sin(3 cos 4x)

Let u = 3 cos 4x and y = 2 sin u

du/dx = 3(-sin 4x)(d/dx)4x

= -12 sin 4x

And dy/du = (d/du) (2 sin u) = 2 cos u

Using the chain rule, we have;

dy/dx = (dy/du) (du/dx)

= (2 cos u) (-12 sin 4x)

= -24 sin 4x cos(3 cos 4x)

Therefore, dy/dx = -24 sin 4x cos(3 cos 4x).

5. Compute the derivative of the function y = arcsin(2x + 1), i.e., y = sin^-1(2x + 1).

Solution:

Given,

y = arcsin(2x + 1)

= sin^-1(2x + 1)

Let y = sin^-1(u) and u = 2x + 1.

dy/du = (d/du) sin^-1(u)

= 1/√(1 – u²)

du/dx = (d/dx)(2x + 1)

= 2

Using the chain rule, we have;

dy/dx = (dy/du)(du/dx)

= [1/√(1 – u²)] (2)

= 2/√[1 – (2x + 1)²]

= 2/√[1 – (4x² + 1 + 4x)]

= 2/√(1 – 4x² – 1 – 4x)

= 2/√(-4x² – 4x)

= 2/2√[-x(x + 1)]

= 1/√[-x(x + 1)]

Hence, dy/dx = 1/√[-x(x + 1)].

6. Using the chain rule, find the derivative of the function y = 1/(x³ + 4x² − 3x − 3)⁶.

Solution:

Given function is:

y = 1/(x³ + 4x² − 3x − 3)⁶

Ley y = 1/t⁶ and t = x³ + 4x² – 3x – 3.

dy/dy = (d/dt)1/t⁶= -6/t⁷ = -6t^-7

dt/dx = (d/dx)(x³ + 4x² – 3x – 3)

= 3x² + 8x – 3

Using the chain rule, we can find dy/dx as:

dy/dx = (-6t^-7) (3x² – 8x – 3)

= -6(3x² – 8x – 3) (x³ + 4x² − 3x − 3)^-7

Thus, dy/dx = -6(3x² – 8x – 3) (x³ + 4x² − 3x − 3)^-7.

7. Find the derivative of y = √[(x + 3)³(x – 1)⁴].

Solution:

Given,

y = √[(x + 3)³(x – 1)⁴]

Let y = √t and t = (x + 3)³(x – 1)⁴

dy/dt = (d/dt)√t = 1/(2√t)

dt/dx = (d/dx)uv, where u = (x + 3)³ and v = (x – 1)⁴

= u(dv/dx) + v(du/dx)

= 4(x + 3)³(x − 1)³ + 3(x − 1)⁴(x + 3)²

= (x + 3)²(x − 1)³(7x + 9)

Using the chain rule, we have;

dy/dx = (dy/dt) (dt/dx)

= [1/(2√t)] [(x + 3)²(x − 1)³(7x + 9)]

= [(x + 3)²(x − 1)³(7x + 9)]/ 2√[(x + 3)³(x – 1)⁴]

= (½) (x – 1)(7x + 9) √(x + 3)

Hence, the derivative of y = √[(x + 3)³(x – 1)⁴] is dy/dx = (½) (x – 1)(7x + 9) √(x + 3).

8. What is the first derivative of the function f(x) = [cos(7x² + 3)]⁴?

Solution:

Given, f(x) = [cos(7x² + 3)]⁴

Let f(x) = y = (cos u)⁴ and u = 7x² + 3

Consider y = (cos u)⁴

Let y = t⁴ and t = cos u

dy/dt = (d/dt)t⁴ = 4t³

dt/du = (d/du) cos u = -sin u

Now, by applying the chain rule, we get;

dy/du = (4t³) (-sin u)

= -4(cos u)³ (sin u)

= -4[cos(7x² + 3)]³ [sin(7x² + 3)]

Consider u = 7x² + 3

du/dx =(d/dx)(7x² + 3) = 14x

Therefore, f’(x) = dy/dx = (dy/du)(du/dx) {using the chain rule}

= -4[cos(7x² + 3)]³ [sin(7x² + 3)] (14x)

= -56x [cos(7x² + 3)]³ [sin(7x² + 3)]

9. Find the derivative of the function y = tan(4x³ – 5x + 2).

Solution:

Given,

y = tan(4x³ – 5x + 2)

Let y = tan u and u = 4x³ – 5x + 2

dy/du = (d/du) tan u

= sec²u

du/dx = (d/dx) (4x³ – 5x + 2)

= 12x² – 5

Thus, by chain rule of differentiation, we get;

dy/dx = (dy/du) (du/dx)

= (sec²u) (12x² – 5)

= (12x² – 5) sec²(4x³ – 5x + 2)

Hence, the derivative of y = tan(4x³ – 5x + 2) is dy/dx = (12x² – 5) sec²(4x³ – 5x + 2).

10. Differentiate the function y = 4 ln(√x) with respect to x.

Solution:

Given,

y = 4 ln(√x)

Let y = 4 ln(u) and u = √x

dy/du = (d/du) 4 ln(u) = 4(1/u) = 4/u

du/dx = (d/dx)(√x) = 1/2√x

Using the chain rule of differentiation, we have;

dy/dx = (dy/du) (du/dx)

= (4/u) (1/2√x)

= (4/2) (1/√x.√x)

= 2/x

Therefore, dy/dx = 2/x.

Practice Questions on Chain Rule

1. Find the derivative of the function y = ln(sin 2x).
2. What is the derivative of y = 3 e^{2x cos x} + 2x?
3. If y = arctan(√cos x), find the derivative of y.
4. Compute the first derivative of the function y = 1/√(2 – x⁴).
5. Using the chain rule, find the derivative of y = ln(e^x + xe^x).