Worksheet · Calculus I · Unit 2A

Chain Rule Worksheet with Answers and Worked Solutions

Twenty-four composition problems from basic powers through nested, product, and quotient combinations.

24 problems60 min estimated timeFoundational to advanced progression

What is included

Build chain-rule fluency with a printable worksheet, complete key, error analysis, and worked HTML solutions.

Skills assessed

composition structure
nested chain rule
product plus chain
quotient plus chain

Prerequisites

basic derivative rules
function composition

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Differentiate $y=(3x+2)^{5}$ .
Differentiate $y=(5x-1)^{4}$ .
Differentiate $y=(2x+4)^{6}$ .
Differentiate $y=(-3x+2)^{3}$ .
Differentiate $y=(7x+1)^{2}$ .
Differentiate $y=(4x-5)^{3}$ .
Differentiate $y=e^{3x^2}$ .
Differentiate $y=e^{\sin x}$ .
Differentiate $y=\ln(5x-2)$ .
Differentiate $y=\ln(x^2+4)$ .
Differentiate $y=\sin(4x^3)$ .
Differentiate $y=\cos(2x-1)$ .
Differentiate $y=\tan(x^2)$ .
Differentiate $y=\sin^3 x$ .
Differentiate $y=(1+(2x-3)^2)^4$ .
Differentiate $y=e^{\sqrt{1+x^2}}$ .
Differentiate $y=\ln(1+\sin^2x)$ .
Differentiate $y=x^2e^{3x}$ .
Differentiate $y=x\sin(x^2)$ .
Differentiate $y=\frac{\ln x}{x^2}$ .
Differentiate $y=\frac{e^x}{1+x^2}$ .
Differentiate $y=(x^2+1)^3(x-2)$ .
Differentiate $y=\frac{\sin(2x)}{x}$ .
Differentiate $y=\sqrt{1+e^{2x}}$ .

Complete worked solutions

Every problem has a source-matched answer and independently reviewed derivation.

01

Problem 1: Differentiate $y=(3x+2)^{5}$ .

Answer: $y'=15(3x+2)^{4}$

Why this method: Basic chain rule matches the mathematical structure before any algebraic cleanup.

Use the power rule on the outer power $u^{5}$ .
Multiply by the inner derivative $3$ .
Therefore the result is $y'=15(3x+2)^{4}$ .

02

Problem 2: Differentiate $y=(5x-1)^{4}$ .

Answer: $y'=20(5x-1)^{3}$

Why this method: Basic chain rule matches the mathematical structure before any algebraic cleanup.

Use the power rule on the outer power $u^{4}$ .
Multiply by the inner derivative $5$ .
Therefore the result is $y'=20(5x-1)^{3}$ .

03

Problem 3: Differentiate $y=(2x+4)^{6}$ .

Answer: $y'=12(2x+4)^{5}$

Why this method: Basic chain rule matches the mathematical structure before any algebraic cleanup.

Use the power rule on the outer power $u^{6}$ .
Multiply by the inner derivative $2$ .
Therefore the result is $y'=12(2x+4)^{5}$ .

04

Problem 4: Differentiate $y=(-3x+2)^{3}$ .

Answer: $y'=-9(-3x+2)^{2}$

Why this method: Basic chain rule matches the mathematical structure before any algebraic cleanup.

Use the power rule on the outer power $u^{3}$ .
Multiply by the inner derivative $-3$ .
Therefore the result is $y'=-9(-3x+2)^{2}$ .

05

Problem 5: Differentiate $y=(7x+1)^{2}$ .

Answer: $y'=14(7x+1)^{1}$

Why this method: Basic chain rule matches the mathematical structure before any algebraic cleanup.

Use the power rule on the outer power $u^{2}$ .
Multiply by the inner derivative $7$ .
Therefore the result is $y'=14(7x+1)^{1}$ .

06

Problem 6: Differentiate $y=(4x-5)^{3}$ .

Answer: $y'=12(4x-5)^{2}$

Why this method: Basic chain rule matches the mathematical structure before any algebraic cleanup.

Use the power rule on the outer power $u^{3}$ .
Multiply by the inner derivative $4$ .
Therefore the result is $y'=12(4x-5)^{2}$ .

07

Problem 7: Differentiate $y=e^{3x^2}$ .

Answer: $y'=6xe^{3x^2}$

Why this method: Exponential or logarithmic chain rule matches the mathematical structure before any algebraic cleanup.

Treat $3x^2$ as the inner function.
Differentiate the outer function and multiply by $6x$ .
Therefore the result is $y'=6xe^{3x^2}$ .

08

Problem 8: Differentiate $y=e^{\sin x}$ .

Answer: $y'=e^{\sin x}\cos x$

Why this method: Exponential or logarithmic chain rule matches the mathematical structure before any algebraic cleanup.

Treat $\sin x$ as the inner function.
Differentiate the outer function and multiply by $\cos x$ .
Therefore the result is $y'=e^{\sin x}\cos x$ .

09

Problem 9: Differentiate $y=\ln(5x-2)$ .

Answer: $y'=\frac5{5x-2}$

Why this method: Exponential or logarithmic chain rule matches the mathematical structure before any algebraic cleanup.

Treat $5x-2$ as the inner function.
Differentiate the outer function and multiply by $5$ .
Therefore the result is $y'=\frac5{5x-2}$ .

10

Problem 10: Differentiate $y=\ln(x^2+4)$ .

Answer: $y'=\frac{2x}{x^2+4}$

Why this method: Exponential or logarithmic chain rule matches the mathematical structure before any algebraic cleanup.

Treat $x^2+4$ as the inner function.
Differentiate the outer function and multiply by $2x$ .
Therefore the result is $y'=\frac{2x}{x^2+4}$ .

11

Problem 11: Differentiate $y=\sin(4x^3)$ .

Answer: $y'=12x^2\cos(4x^3)$

Why this method: Trigonometric chain rule matches the mathematical structure before any algebraic cleanup.

Identify the outer trigonometric function and preserve its inner input.
Differentiate the inner expression and multiply.
Therefore the result is $y'=12x^2\cos(4x^3)$ .

12

Problem 12: Differentiate $y=\cos(2x-1)$ .

Answer: $y'=-2\sin(2x-1)$

Why this method: Trigonometric chain rule matches the mathematical structure before any algebraic cleanup.

Identify the outer trigonometric function and preserve its inner input.
Differentiate the inner expression and multiply.
Therefore the result is $y'=-2\sin(2x-1)$ .

13

Problem 13: Differentiate $y=\tan(x^2)$ .

Answer: $y'=2x\sec^2(x^2)$

Why this method: Trigonometric chain rule matches the mathematical structure before any algebraic cleanup.

Identify the outer trigonometric function and preserve its inner input.
Differentiate the inner expression and multiply.
Therefore the result is $y'=2x\sec^2(x^2)$ .

14

Problem 14: Differentiate $y=\sin^3 x$ .

Answer: $y'=3\sin^2x\cos x$

Why this method: Trigonometric chain rule matches the mathematical structure before any algebraic cleanup.

Identify the outer trigonometric function and preserve its inner input.
Differentiate the inner expression and multiply.
Therefore the result is $y'=3\sin^2x\cos x$ .

15

Problem 15: Differentiate $y=(1+(2x-3)^2)^4$ .

Answer: $y'=16(2x-3)(1+(2x-3)^2)^3$

Why this method: Nested chain rule matches the mathematical structure before any algebraic cleanup.

Write the composition as nested layers from outside to inside.
Differentiate each layer once and multiply the factors.
Therefore the result is $y'=16(2x-3)(1+(2x-3)^2)^3$ .

16

Problem 16: Differentiate $y=e^{\sqrt{1+x^2}}$ .

Answer: $y'=\frac{x e^{\sqrt{1+x^2}}}{\sqrt{1+x^2}}$

Why this method: Nested chain rule matches the mathematical structure before any algebraic cleanup.

Write the composition as nested layers from outside to inside.
Differentiate each layer once and multiply the factors.
Therefore the result is $y'=\frac{x e^{\sqrt{1+x^2}}}{\sqrt{1+x^2}}$ .

17

Problem 17: Differentiate $y=\ln(1+\sin^2x)$ .

Answer: $y'=\frac{2\sin x\cos x}{1+\sin^2x}$

Why this method: Nested chain rule matches the mathematical structure before any algebraic cleanup.

Write the composition as nested layers from outside to inside.
Differentiate each layer once and multiply the factors.
Therefore the result is $y'=\frac{2\sin x\cos x}{1+\sin^2x}$ .

18

Problem 18: Differentiate $y=x^2e^{3x}$ .

Answer: $y'=e^{3x}(2x+3x^2)$

Why this method: product plus chain matches the mathematical structure before any algebraic cleanup.

Apply the product plus chain structure before simplifying.
Retain every factor contributed by an inner derivative.
A factored final form is $e^{3x}(2x+3x^2)$ .
Therefore the result is $y'=e^{3x}(2x+3x^2)$ .

19

Problem 19: Differentiate $y=x\sin(x^2)$ .

Answer: $y'=\sin(x^2)+2x^2\cos(x^2)$

Why this method: product plus chain matches the mathematical structure before any algebraic cleanup.

Apply the product plus chain structure before simplifying.
Retain every factor contributed by an inner derivative.
A factored final form is $\sin(x^2)+2x^2\cos(x^2)$ .
Therefore the result is $y'=\sin(x^2)+2x^2\cos(x^2)$ .

20

Problem 20: Differentiate $y=\frac{\ln x}{x^2}$ .

Answer: $y'=\frac{1-2\ln x}{x^3}$

Why this method: quotient plus chain matches the mathematical structure before any algebraic cleanup.

Apply the quotient plus chain structure before simplifying.
Retain every factor contributed by an inner derivative.
A factored final form is $\frac{1-2\ln x}{x^3}$ .
Therefore the result is $y'=\frac{1-2\ln x}{x^3}$ .

21

Problem 21: Differentiate $y=\frac{e^x}{1+x^2}$ .

Answer: $y'=\frac{e^x(1+x^2-2x)}{(1+x^2)^2}$

Why this method: quotient rule matches the mathematical structure before any algebraic cleanup.

Apply the quotient rule structure before simplifying.
Retain every factor contributed by an inner derivative.
A factored final form is $\frac{e^x(1+x^2-2x)}{(1+x^2)^2}$ .
Therefore the result is $y'=\frac{e^x(1+x^2-2x)}{(1+x^2)^2}$ .

22

Problem 22: Differentiate $y=(x^2+1)^3(x-2)$ .

Answer: $y'=(x^2+1)^2(7x^2-12x+1)$

Why this method: product plus chain matches the mathematical structure before any algebraic cleanup.

Apply the product plus chain structure before simplifying.
Retain every factor contributed by an inner derivative.
A factored final form is $(x^2+1)^2(7x^2-12x+1)$ .
Therefore the result is $y'=(x^2+1)^2(7x^2-12x+1)$ .

23

Problem 23: Differentiate $y=\frac{\sin(2x)}{x}$ .

Answer: $y'=\frac{2x\cos(2x)-\sin(2x)}{x^2}$

Why this method: quotient plus chain matches the mathematical structure before any algebraic cleanup.

Apply the quotient plus chain structure before simplifying.
Retain every factor contributed by an inner derivative.
A factored final form is $\frac{2x\cos(2x)-\sin(2x)}{x^2}$ .
Therefore the result is $y'=\frac{2x\cos(2x)-\sin(2x)}{x^2}$ .

24

Problem 24: Differentiate $y=\sqrt{1+e^{2x}}$ .

Answer: $y'=\frac{e^{2x}}{\sqrt{1+e^{2x}}}$

Why this method: nested chain matches the mathematical structure before any algebraic cleanup.

Apply the nested chain structure before simplifying.
Retain every factor contributed by an inner derivative.
A factored final form is $\frac{e^{2x}}{\sqrt{1+e^{2x}}}$ .
Therefore the result is $y'=\frac{e^{2x}}{\sqrt{1+e^{2x}}}$ .

Common errors

Forgetting an inner derivative.
Applying the chain rule to a sum or product as a single composition.
Simplifying before the rule structure is secure.