AP Calculus AB Power, Sum and Basic Function Derivative Rules Worksheet Solutions (PDF)

AP Calculus AB math worksheet solutions: Power, Sum and Basic Function Derivative Rules – Maths PDF

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This derivative rules answers for AP Calculus AB covers power, sum and basic function derivative rules (Common Core AP CALC AB 2.5, AP CALC AB 2.6, AP CALC AB 2.7), with a free PDF to print.

This answer key gives a full worked solution for every problem in the derivative rules worksheet. Compare each step, not only the final answer, because AP graders award credit for a correct setup and clear justification. Final answers are in bold. Redo any problem you missed before moving on to the test.

AP Calculus AB · Common Core State Standards: AP CALC AB 2.5, AP CALC AB 2.6, AP CALC AB 2.7

Derivative rules answers: exercises 1 to 10

Exercise 1 – Constant and power

The derivative of a constant is $0$.
$g^{\prime}(x)=5x^{4}$. $5x^{4}$

Exercise 2 – Higher power

Apply the power rule: bring down $8$ and subtract $1$ from the exponent. Final answer: $8 x^{7}$.

Exercise 3 – Polynomial

Differentiate term by term: $f^{\prime}(x)=12 x^{2} – 10 x + 7$.

Final answer: $12 x^{2} – 10 x + 7$.

Exercise 4 – Polynomial with a fraction

Differentiate term by term: $f^{\prime}(x)=12 x^{3} + x$.

Final answer: $12 x^{3} + x$.

Exercise 5 – Root and reciprocal

Rewrite with powers: $\sqrt{x} + \frac{5}{x^{2}}$.

Differentiate term by term: $f^{\prime}(x)=- \frac{10}{x^{3}} + \frac{1}{2 \sqrt{x}}$.

Final answer: $- \frac{10}{x^{3}} + \frac{1}{2 \sqrt{x}}$.

Exercise 6 – Simplify first

Divide each term by $x$: $f(x)=x^{4}-3x$.

Then $f^{\prime}(x)=4x^{3}-3$. Final answer: $4x^{3}-3$.

Exercise 7 – Negative and fractional powers

Rewrite: $f(x)=6x^{-3}-2x^{1/3}$.

$f^{\prime}(x)=-18x^{-4}-\frac{2}{3}x^{-2/3}=-\frac{18}{x^{4}}-\frac{2}{3x^{2/3}}$.

Final answer: $-\frac{18}{x^{4}}-\frac{2}{3\sqrt[3]{x^{2}}}$.

Exercise 8 – Sine and cosine

The term $\pi x$ has derivative $\pi$ because $\pi$ is a constant.

$f^{\prime}(x)=4(-\sin x)+\frac{1}{2}\cos x+\pi$.

Final answer: $-4\sin x+\frac{1}{2}\cos x+\pi$.

Exercise 9 – Exponential and logarithm

$f^{\prime}(x)=5e^{x}-4\cdot\frac{1}{x}$.

Final answer: $5e^{x}-\frac{4}{x}$.

Exercise 10 – Mixed functions

The term $\pi$ is a constant, so its derivative is $0$.

$f^{\prime}(x)=2x+2\cos x-e^{x}$. Final answer: $2x+2\cos x-e^{x}$.

Exercises 11 and up

Exercise 11 – Evaluate a derivative

$f^{\prime}(x)=3x^{2}-4x+1$.

$f^{\prime}(2)=12-8+1=5$. Final answer: $5$.

Exercise 12 – Horizontal tangents of a quartic

$f^{\prime}(x)=4x^{3}-16x=4x(x-2)(x+2)$. It is zero at $x=0,\pm 2$.

$f(0)=3$ and $f(\pm 2)=16-32+3=-13$.

Final answer: $(0,3),\ (2,-13),\ (-2,-13)$.

Exercise 13 – Tangent line to a parabola

$f(4)=16-24+5=-3$ and $f^{\prime}(x)=2x-6$, so $f^{\prime}(4)=2$.

Point-slope: $y+3=2(x-4)$, so $y=2x-11$.

Final answer: $y=2x-11$.

The tangent line at x = 4.

Exercise 14 – Tangent line to a cubic

$f(-1)=-1$ and $f^{\prime}(-1)=3(-1)^{2}=3$.

$y+1=3(x+1)$, so $y=3x+2$.

The tangent to y = x³ at x = −1.

Exercise 15 – Normal line

$f(4)=2$ and $f^{\prime}(x)=\frac{1}{2\sqrt{x}}$, so $f^{\prime}(4)=\frac{1}{4}$.

The normal slope is the negative reciprocal: $-4$. So $y-2=-4(x-4)$, which is $y=-4x+18$.

Final answer: $y=-4x+18$.

Exercise 16 – Parallel tangent lines

Parallel lines have equal slopes, so solve $f^{\prime}(x)=9$: $3x^{2}-3=9$, so $x^{2}=4$ and $x=\pm 2$.

$f(2)=8-6=2$ and $f(-2)=-8+6=-2$.

Final answer: $(2,2)$ and $(-2,-2)$.

Two parallel tangents of slope 9.

Exercise 17 – Tangent to the exponential

$f(0)=1$ and $f^{\prime}(0)=e^{0}=1$.

Final answer: $y=x+1$.

Exercise 18 – Tangent to the logarithm

$f(e)=1$ and $f^{\prime}(e)=\frac{1}{e}$.

$y-1=\frac{1}{e}(x-e)$ simplifies to $y=\frac{x}{e}$.

Final answer: $y=\frac{x}{e}$.

Exercise 19 – Velocity of a particle

$v(t)=s^{\prime}(t)=3t^{2}-12t+9=3(t-1)(t-3)$.

At rest when $v(t)=0$: $t=1$ or $t=3$.

Final answer: $v(t)=3t^{2}-12t+9\text{ ft/s}$; at rest at $t=1$ s and $t=3$ s.

Velocity is zero at t = 1 and t = 3.

Exercise 20 – Maximum profit

$P^{\prime}(x)=-2x+50$. Set it equal to $0$: $x=25$.

$P(25)=-625+1250-200=425$.

Final answer: horizontal tangent at $x=25$ candles, where the profit is \$425.

Exercise 21 – Finding coefficients

Point: $f(1)=a+b=3$. Slope: $f^{\prime}(x)=2ax+b$, so $f^{\prime}(1)=2a+b=2$.

Subtract the first equation from the second: $a=-1$, then $b=4$.

Final answer: $a=-1,\ b=4$.

Exercise 22 – Tangent through an outside point

The slope is $2a$, so the tangent is $y-a^{2}=2a(x-a)$. At $x=0$: $y=a^{2}-2a^{2}=-a^{2}$.

Set $-a^{2}=-4$: $a=\pm 2$.

Final answer: $a=2$ or $a=-2$.

Key facts behind this derivative rules answers

Key facts

Power rule: d/dx of x^n equals n x^(n-1); the derivative of a constant is 0.
Rewrite roots and reciprocals as powers before differentiating.
The derivatives of sin x, cos x, e^x and ln x are cos x, -sin x, e^x and 1/x.
A normal line has slope equal to the negative reciprocal of the tangent slope.

How to use this derivative rules answers

Cover each solution, solve the exercise again, then uncover it and compare every line. If your answer differs, find the first step that changed and redo the exercise from there.

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