Chain Rule Practice Problems with Answers and Solutions
12 chain rule practice problems with answers and full step-by-step solutions, from (3x+1)⁵ to nested functions like sin²(4x) and e^(cos x).
Derivatives Updated 3 min read
At a glance
- Problems
- 12, with answers and full solutions
- Level
- Calculus 1 and AP Calculus AB, Unit 3
- You need
- The power rule and the basic trig, exponential and log derivatives
- Main idea
- Differentiate the outside, keep the inside, then multiply by the derivative of the inside
- Practice with new numbers
- Chain rule lessons
On this page
These chain rule practice problems start with a single layer and build up to nested functions and combinations with the product and quotient rules. Try each one before you read the solution. The answers are listed first so you can check quickly, and the full solutions follow.
The rule you need:
Differentiate the outside function, leave the inside alone, then multiply by the derivative of the inside. For a refresher, read the chain rule page first.
Problems
One layer
- y = (3x + 1)^5
- y = \sqrt{x^2 + 9}
- y = \sin(4x)
- y = e^{x^2}
- y = \ln(x^3 + 2x)
- y = \dfrac{1}{(2x - 5)^3}
Two or more layers
- y = \sin^2(4x)
- y = e^{\cos x}
- y = \tan\!\left(\sqrt{x}\right)
With the product or quotient rule
- y = x^2 \sin(3x)
- y = x\,e^{-2x}
- y = \dfrac{(x+1)^4}{x^2}
Answers
| Problem | Answer |
|---|---|
| 1 | 15(3x + 1)^4 |
| 2 | \dfrac{x}{\sqrt{x^2 + 9}} |
| 3 | 4\cos(4x) |
| 4 | 2x\,e^{x^2} |
| 5 | \dfrac{3x^2 + 2}{x^3 + 2x} |
| 6 | -\dfrac{6}{(2x - 5)^4} |
| 7 | 8\sin(4x)\cos(4x) = 4\sin(8x) |
| 8 | -\sin x\,e^{\cos x} |
| 9 | \dfrac{\sec^2\!\left(\sqrt{x}\right)}{2\sqrt{x}} |
| 10 | 2x\sin(3x) + 3x^2\cos(3x) |
| 11 | (1 - 2x)\,e^{-2x} |
| 12 | \dfrac{2(x+1)^3(x - 1)}{x^3} |
Step-by-step solutions
1. The inside is u = 3x + 1 with u' = 3. The outside is u^5, with derivative 5u^4:
2. Write y = (x^2 + 9)^{1/2}. The inside derivative is 2x:
3. The outside is \sin u, whose derivative is \cos u, and u' = 4:
4. The derivative of e^u is e^u \cdot u', and u = x^2:
5. The derivative of \ln u is \frac{u'}{u} (see the derivative of ln x):
6. Rewrite as y = (2x - 5)^{-3}, then use the power rule on the outside:
7. There are three layers: a square, a sine and 4x. Work from the outside in:
With the double-angle identity 2\sin\theta\cos\theta = \sin 2\theta, this is 4\sin(8x).
8. The outside is e^u with u = \cos x and u' = -\sin x:
9. The outside is \tan u with derivative \sec^2 u, and u = \sqrt{x} has u' = \frac{1}{2\sqrt{x}}:
10. Use the product rule, with the chain rule for \sin(3x):
11. Product rule again. The derivative of e^{-2x} is -2e^{-2x}:
12. Use the quotient rule, with the chain rule on the top:
Factor 2x(x+1)^3 out of the top: 2x(x+1)^3\big(2x - (x + 1)\big) = 2x(x+1)^3(x-1). Cancel one x:
Common mistakes on these problems
- Dropping the inside derivative. In problem 3, \cos(4x) alone is wrong. The factor 4 is the whole point of the chain rule.
- Changing the inside. The derivative of \sin(4x) is 4\cos(4x), not 4\cos(4) or \cos(4). The inside stays exactly as it was.
- Missing a layer. Problem 7 has three layers, so the answer has three factors. Count them before you finish.
- Expanding too early. In problem 1, expanding (3x+1)^5 first is possible but slow and error-prone. The chain rule takes one line.
Keep practicing
These 12 problems use fixed numbers. For unlimited practice with new numbers every time and instant feedback on each step, try the chain rule lessons, then the mixed chain rule practice. For the theory with more examples, read the chain rule guide.
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