AP Calculus AB The Chain Rule Test Solutions and Scoring (PDF)
This chain rule test answers for AP Calculus AB covers the chain rule (Common Core AP CALC AB 3.1), with a free PDF to print.
Use this answer key to score the chain rule test out of 20 points. Each solution shows the work and a scoring guide that says how the points are awarded. Give partial credit for a correct setup, a correct derivative or integral, and a correct final statement with units where they apply.
AP Calculus AB · Common Core State Standards: AP CALC AB 3.1
Chain rule test answers: worked solutions
Question 1 – Power chain [2 pts]
\(f^{\prime}(x)=6(4x-7)^{5}\cdot 4\). Final answer: \(24(4x-7)^{5}\).
Scoring: 1 point for the outer derivative; 1 point for the inner factor 4.
Question 2 – Trig chain [2 pts]
\(g^{\prime}(x)=\cos(x^{3})\cdot 3x^{2}-\sin(5x)\cdot 5\).
Final answer: \(3x^{2}\cos(x^{3})-5\sin(5x)\).
Scoring: 1 point per term.
Question 3 – Exponential and log chain [2 pts]
\(h^{\prime}(x)=e^{x^{2}-3x}(2x-3)+\frac{4}{4x+1}\).
Final answer: \((2x-3)e^{x^{2}-3x}+\frac{4}{4x+1}\).
Scoring: 1 point per term.
Question 4 – Three layers [3 pts]
Layers: cube, sine, then \(x^{2}+1\).
\(k^{\prime}(x)=3\sin^{2}(x^{2}+1)\cdot\cos(x^{2}+1)\cdot 2x\).
Final answer: \(6x\sin^{2}(x^{2}+1)\cos(x^{2}+1)\).
Scoring: 1 point for each layer correctly differentiated.
Question 5 – Tangent line [3 pts]
\(y(1)=\sqrt{3}\). \(y^{\prime}=\frac{4x}{2\sqrt{2x^{2}+1}}=\frac{2x}{\sqrt{2x^{2}+1}}\), so \(y^{\prime}(1)=\frac{2}{\sqrt{3}}\).
Final answer: \(y-\sqrt{3}=\frac{2}{\sqrt{3}}(x-1)\).
Scoring: 1 point for y(1); 1 point for y′(1); 1 point for the equation.
Question 6 – Composite from a table [3 pts]
- \(f^{\prime}(g(1))\cdot g^{\prime}(1)=f^{\prime}(2)\cdot(-5)=3(-5)=-15\).
- \(p^{\prime}(4)=2f(4)f^{\prime}(4)=2(7)(6)=84\).
Scoring: (a) 1 point for g(1) = 2 and 1 point for the product; (b) 1 point.
Question 7 – Product and chain [2 pts]
\(p^{\prime}(x)=1\cdot e^{3x}+x\cdot 3e^{3x}=e^{3x}(1+3x)\).
Final answer: \(e^{3x}(1+3x)\).
Scoring: 1 point for the product rule; 1 point for the chain factor 3.
Question 8 – Tide height [3 pts]
- \(h^{\prime}(t)=4\cos\left(\frac{\pi t}{6}\right)\cdot\frac{\pi}{6}=\frac{2\pi}{3}\cos\left(\frac{\pi t}{6}\right)\).
- \(h^{\prime}(0)=\frac{2\pi}{3}\approx 2.09\) ft/h and \(h^{\prime}(3)=\frac{2\pi}{3}\cos\frac{\pi}{2}=0\).
- At 3 hours the tide is at its highest point (16 feet), so it is momentarily neither rising nor falling.
Scoring: 1 point per part.
Total score
20 points
Key facts behind this chain rule test answers
- Chain rule: dy/dx = f′(g(x)) · g′(x), or dy/du times du/dx.
- Outer derivative first, inner function unchanged, then multiply by the inner derivative.
- d/dx e^(g(x)) = e^(g(x)) g′(x) and d/dx ln(g(x)) = g′(x)/g(x).
- Functions with three layers need the chain rule twice.
How to use this chain rule test answers
Score your test with the point values shown for each question. Then read the worked solution of every question where you lost points and note the method to review.
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