Differentiation: composite, implicit and inverse functions practice problems AP Calculus, Unit 3
The chain rule, implicit differentiation, derivatives of inverse functions and inverse trigonometric functions. Try each question before you look: use the hint if you are stuck, then reveal the worked solution one step at a time.
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Key equations for differentiation: composite, implicit and inverse functions
| d/dx f(g(x)) = f'(g(x)) g'(x) | chain rule |
|---|---|
| (f−1)'(b) = 1 / f'(a), where f(a) = b | derivative of an inverse |
| d/dx arctan x = 1/(1 + x2) | inverse tangent |
| d/dx ln(u) = u'/u | logs of a function |
Questions marked No calculator should be done by hand, as in Part A of the multiple-choice section. You can type fractions, such as 3/4. The full sheet is in our AP Calculus formula sheet.
Warm-up questions
Quick checks on the core ideas. If these feel easy, go straight to the exam-style questions.
Hint
Outside function: u4. Inside: 3x + 1.
Worked solution
- Chain rule: 4(3x + 1)3 × 3.
- At x = 0: 4 × 1 × 3 = 12.
Hint
The derivative of eu is eu u'.
Worked solution
- d/dx e2x = 2e2x.
- At x = 0: 2e0 = 2.
Hint
d/dx ln u = u'/u.
Worked solution
- u = x2 + 1, so u' = 2x.
- The derivative is 2x/(x2 + 1). The answer is B.
Exam-style questions
Questions at the level of the AP Calculus exam. The ones marked Challenge combine more than one idea.
Hint
Differentiate both sides with respect to x, treating y as a function of x.
Worked solution
- 2x + 2y(dy/dx) = 0, so dy/dx = −x/y.
- At (3, 4): dy/dx = −3/4 = −0.75.
Hint
Use (f−1)'(b) = 1/f'(a) where f(a) = b.
Worked solution
- f(1) = 2, so a = 1.
- f'(x) = 3x2 + 1, so f'(1) = 4.
- (f−1)'(2) = 1/4.
Hint
Chain rule with the arctan derivative.
Worked solution
- d/dx arctan(2x) = 2/(1 + 4x2).
- At x = 0: 2.
Hint
Chain rule: 3(inside)2 × (derivative of inside).
Worked solution
- dy/dx = 3(x2 + 1)2 × 2x.
- At x = 1: 3 × 4 × 2 = 24.
Hint
Differentiate implicitly. The xy term needs the product rule.
Worked solution
- 2x + (y + x dy/dx) + 2y dy/dx = 0.
- dy/dx = −(2x + y)/(x + 2y).
- At (1, 2): −4/5 = −0.8.
Free-response practice
One original question in the style of the exam's free-response section. Answer it in full, with working, before you open the scoring guide.
No-calculator free-response style · 6 points · about 15 minutes
Consider the curve x2 + xy + y2 = 7.
- (a) Show that dy/dx = −(2x + y)/(x + 2y).
- (b) Write the equation of the tangent line at (1, 2).
- (c) Find the x-coordinates of the points where the tangent is horizontal.
Write your full answer on paper first, then open the scoring guide and tick each point you earned.
Scoring guide and model answer
Part (a)
Differentiating: 2x + y + x dy/dx + 2y dy/dx = 0, so (x + 2y) dy/dx = −(2x + y).
Part (b)
Slope = −4/5, so y − 2 = −(4/5)(x − 1).
Part (c)
dy/dx = 0 when y = −2x. Substituting: x2 − 2x2 + 4x2 = 7, so 3x2 = 7 and x = ±√(7/3).
Your mark: 0 of 6
Common mistakes in differentiation: composite, implicit and inverse functions
- Forgetting to multiply by the derivative of the inside function.
- Dropping dy/dx on a y term when differentiating implicitly.
- Using f'(b) instead of f'(a) for the derivative of an inverse.
Keep going
Revise the ideas behind these questions:
Then try real exam questions: AP Calculus past papers, with official scoring guidelines. · All AP Calculus units