AP Calculus AB Derivatives of Inverse Functions Test Solutions and Scoring (PDF)

AP Calculus AB math test solutions: Derivatives of Inverse Functions – Maths PDF

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This inverse derivatives test answers for AP Calculus AB covers derivatives of inverse functions (Common Core AP CALC AB 3.3, AP CALC AB 3.4, AP CALC AB 3.5, AP CALC AB 3.6), with a free PDF to print.

Use this answer key to score the inverse derivatives 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 3.3, AP CALC AB 3.4, AP CALC AB 3.5, AP CALC AB 3.6

Inverse derivatives test answers: worked solutions

Question 1 – Inverse from a table [3 pts]

\(f(2)=9\), so \(\frac{1}{f^{\prime}(2)}=\frac{1}{5}\).
\(f(3)=15\), so \(\frac{1}{f^{\prime}(3)}=\frac{1}{8}\).
\(f(1)=4\), so \(f^{-1}(4)=1\).

Scoring: 1 point per part.

Question 2 – Inverse of a cubic [2 pts]

\(f(1)=0\), so \(a=1\). \(f^{\prime}(x)=3x^{2}+2\), so \(f^{\prime}(1)=5\).

Final answer: \(\frac{1}{5}\).

Graph of f(x) = x³ + 2x − 3.

Scoring: 1 point for finding a and f′(a); 1 point for the reciprocal.

Question 3 – Inverse trig derivatives [2 pts]

\(g^{\prime}(x)=\frac{1}{\sqrt{1-x^{2}}}+\frac{2}{1+4x^{2}}\).

Scoring: 1 point per term.

Question 4 – Inverse trig and a root [2 pts]

\(h^{\prime}(x)=\frac{1}{1+(\sqrt{x})^{2}}\cdot\frac{1}{2\sqrt{x}}=\frac{1}{2\sqrt{x}(1+x)}\).

Scoring: 1 point for the arctangent derivative; 1 point for the chain factor.

Question 5 – Second derivative [3 pts]

\(f^{\prime}(x)=5x^{4}-6x^{2}+\cos x\) and \(f^{\prime\prime}(x)=20x^{3}-12x-\sin x\).

\(f^{\prime\prime}(0)=0-0-0=0\).

Scoring: 1 point for f′; 1 point for f″; 1 point for the value.

Question 6 – Acceleration [2 pts]

\(v=6t^{2}-30t+36=6(t-2)(t-3)\), zero at \(t=2\) and \(t=3\). \(a=12t-30\), so \(a(3)=6\text{ m/s}^{2}\).

Scoring: 1 point for the velocity zeros; 1 point for the acceleration.

Question 7 – Choosing rules [3 pts]

Product rule: \(e^{x}\cos x-e^{x}\sin x=e^{x}(\cos x-\sin x)\).
Quotient rule: \(\frac{\frac{1}{x}\cdot x^{2}-\ln x\cdot 2x}{x^{4}}=\frac{1-2\ln x}{x^{3}}\).
Chain rule: \(20(5x-1)^{3}\).

Scoring: 1 point per part (rule and result).

Question 8 – Inverse in context [3 pts]

\(f(2)=1\) and \(f^{\prime}(x)=\frac{3x^{2}}{8}\), so \(f^{\prime}(2)=\frac{3}{2}\).
\(g^{\prime}(1)=\frac{1}{f^{\prime}(2)}=\frac{2}{3}\).
\(g(1)=2\), so \(y-2=\frac{2}{3}(x-1)\).

Graphs of f and g = f⁻¹.

Scoring: 1 point per part.

Total score

Total

20 points

Key facts behind this inverse derivatives test answers

Key facts

(f⁻¹)′(b) = 1/f′(a) where f(a) = b; evaluate f′ at a, not at b.
d/dx sin⁻¹x = 1/√(1-x²), d/dx cos⁻¹x = -1/√(1-x²), d/dx tan⁻¹x = 1/(1+x²).
The second derivative is the derivative of the first derivative; acceleration is the second derivative of position.
Choose the rule by the outermost operation: sum, product, quotient or composition.

How to use this inverse derivatives 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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