Pattern recognition is what makes chain rule problems feel easy or hard. The mechanics are always the same — outer derivative times inner derivative — but you need to quickly identify what the "outer" and "inner" functions are. These problems build that recognition. Do each one before looking at the solution.

Quick Formula Reminder

If y = f(g(x)), then dy/dx = f'(g(x)) · g'(x) Outer derivative × inner derivative

Basic Problems (1–7)

📋 Problem 1 — y = (3x + 2)⁵
Outer5u⁴ = 5(3x+2)⁴
Inner3x+2 → 3
Answer15(3x + 2)⁴
📋 Problem 2 — y = √(x² + 1)
Rewrite(x²+1)^(1/2)
Outer½(x²+1)^(−½)
Inner2x
Answerx / √(x² + 1)
📋 Problem 3 — y = e^(4x)
Outere^(4x)
Inner4
Answer4e^(4x)
📋 Problem 4 — y = sin(2x + π)
Outercos(2x + π)
Inner2
Answer2cos(2x + π)
📋 Problem 5 — y = ln(5x)
Outer1/(5x)
Inner5
Answer1/x (the 5s cancel)
📋 Problem 6 — y = cos(x³)
Answer−3x²sin(x³)
📋 Problem 7 — y = (x² − 3x)⁴
Outer4(x²−3x)³
Inner2x − 3
Answer4(2x−3)(x²−3x)³

Intermediate Problems (8–14)

📋 Problem 8 — y = e^(sin x)
Answercos(x)·e^(sin x)
📋 Problem 9 — y = sin²(x)
Outer2sin(x)
Innercos(x)
Answer2sin(x)cos(x) = sin(2x)
📋 Problem 10 — y = tan(x² + 1)
Outersec²(x²+1)
Inner2x
Answer2x·sec²(x²+1)
📋 Problem 11 — y = ln(sin x)
Outer1/sin(x)
Innercos(x)
Answercot(x)
📋 Problem 12 — y = (e^x + 1)³
Outer3(eˣ+1)²
Innereˣ
Answer3eˣ(eˣ+1)²
📋 Problem 13 — y = 1/(3x−1)² = (3x−1)^(−2)
Outer−2(3x−1)^(−3)
Inner3
Answer−6/(3x−1)³
📋 Problem 14 — y = e^(x² − 3x)
Answer(2x−3)·e^(x²−3x)

Advanced Problems (15–20)

📋 Problem 15 — y = sin(cos(2x)) [Double chain rule]
Layer 1d/dx[sin(u)] = cos(u) = cos(cos(2x))
Layer 2d/dx[cos(2x)] = −2sin(2x)
Answer−2sin(2x)·cos(cos(2x))
📋 Problem 16 — y = √(e^(2x) + 1)
Outer1/(2√(e^(2x)+1))
Inner2e^(2x)
Answere^(2x) / √(e^(2x)+1)
📋 Problem 17 — y = ln(x² + sin x)
Answer(2x + cos x) / (x² + sin x)
📋 Problem 18 — y = [ln(x²+1)]³
Layer 13[ln(x²+1)]²
Layer 2d/dx[ln(x²+1)] = 2x/(x²+1)
Answer6x[ln(x²+1)]² / (x²+1)
📋 Problem 19 — y = e^(sin(3x)) [Triple chain rule]
Layer 1e^(sin(3x))
Layer 2cos(3x)
Layer 33
Answer3cos(3x)·e^(sin(3x))
📋 Problem 20 — y = tan²(e^x) [Triple chain rule]
Layer 12tan(eˣ)
Layer 2sec²(eˣ)
Layer 3eˣ
Answer2eˣ·tan(eˣ)·sec²(eˣ)

Frequently Asked Questions

How do I know when to use the chain rule?▾
Use the chain rule whenever you have a function inside another function — a composite. Ask: "Is there something more complex than just x inside the outer function?" If yes, chain rule applies.
What is the most common chain rule mistake?▾
Forgetting to multiply by the inner derivative. Students write d/dx[sin(x²)] = cos(x²) when the correct answer is 2x·cos(x²). Always ask what's inside and differentiate it too.

How to Use This Practice Set

Problems 1–7 are straightforward single applications. Problems 8–14 require combining the chain rule with product or quotient rule. Problems 15–20 involve three or more layers. If you can do problems 15–20 reliably without looking anything up, your chain rule is solid.

The single most important habit: before writing a single symbol, identify the outermost function. Is it a power? A trig function? An exponential? A log? Once you name the outer function, the rest follows mechanically.

The 5 Most Common Chain Rule Patterns

Power of a function: d/dx[un] = nun−1u' Trig of a function: d/dx[sin(u)] = cos(u)·u' Exp of a function: d/dx[eu] = eu·u' Log of a function: d/dx[ln(u)] = u'/u Root of a function: d/dx[√u] = u'/(2√u)

Every chain rule problem fits one of these five patterns (or a combination). Identify which pattern applies before computing.

Problems Requiring Both Chain and Product

📋 d/dx[eˣ·ln(x²+1)]
Product rule= eˣ·ln(x²+1) + eˣ·d/dx[ln(x²+1)]
Chain on ln= eˣ·ln(x²+1) + eˣ·(2x/(x²+1))
Factor eˣ= eˣ[ln(x²+1) + 2x/(x²+1)]
📋 d/dx[sin³(x)·cos(x)]
Product rule= d/dx[sin³x]·cos x + sin³x·d/dx[cos x]
Chain on sin³= 3sin²x·cos x·cos x + sin³x·(−sin x)
Simplify= 3sin²x·cos²x − sin⁴x

How to Check Your Answers

For each derivative you compute, verify it makes dimensional and sign sense: if f(x) is increasing at x=1, then f'(1) should be positive. If f(x) has a local maximum at x=2, then f'(2) = 0. For a numerical check: compute f(x+0.001) and f(x) and estimate [f(x+0.001) − f(x)]/0.001. This should closely match your derivative formula evaluated at x.

AM
Dr. Aisha Malik, PhD Mathematics
Senior Lecturer in Applied Mathematics · 12 years teaching calculus
Dr. Malik holds a PhD in Applied Mathematics from the University of Edinburgh and has taught calculus to over 4,000 students. She has reviewed this article for mathematical accuracy and pedagogical clarity.
Technically reviewed by: Prof. James Chen, Stanford Mathematics Department · April 2026