AP Calculus AB The Derivative as a Limit Worksheet Solutions (PDF)

AP Calculus AB math worksheet solutions: The Derivative as a Limit – Maths PDF

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This derivative definition answers for AP Calculus AB covers the derivative as a limit (Common Core AP CALC AB 2.1, AP CALC AB 2.2, AP CALC AB 2.3, AP CALC AB 2.4), with a free PDF to print.

This answer key gives a full worked solution for every problem in the derivative definition 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 2.1, AP CALC AB 2.2, AP CALC AB 2.3, AP CALC AB 2.4

Derivative definition answers: exercises 1 to 10

Exercise 1 – Average rate of change

$f(1)=2$ and $f(4)=17$. The average rate is $\frac{17-2}{4-1}=\frac{15}{3}=5$.

Final answer: $5$.

Exercise 2 – Secant slope from a table

$\frac{9.5-7}{0.5}=5$ and $\frac{13-9.5}{0.5}=7$.

Final answer: $5$ and $7$; the interval $[2.5,3]$ has the greater average rate.

Exercise 3 – Derivative at a point, x squared

$\frac{(3+h)^{2}-9}{h}=\frac{6h+h^{2}}{h}=6+h$.

As $h\to 0$, $6+h\to 6$. Final answer: $f^{\prime}(3)=6$.

Exercise 4 – Derivative at a point, a quadratic

$f(1)=1$ and $f(1+h)=3(1+2h+h^{2})-2-2h=1+4h+3h^{2}$.

$\frac{4h+3h^{2}}{h}=4+3h\to 4$. Final answer: $f^{\prime}(1)=4$.

Exercise 5 – Derivative of a reciprocal

$\frac{\frac{1}{2+h}-\frac{1}{2}}{h}=\frac{2-(2+h)}{2(2+h)h}=\frac{-h}{2(2+h)h}=\frac{-1}{2(2+h)}$.

As $h\to 0$: $\frac{-1}{2\cdot 2}=-\frac{1}{4}$. Final answer: $f^{\prime}(2)=-\frac{1}{4}$.

Exercise 6 – Derivative of a square root

$\frac{\sqrt{4+h}-2}{h}$ gives $\frac{0}{0}$. Multiply by the conjugate: $\frac{(4+h)-4}{h(\sqrt{4+h}+2)}=\frac{1}{\sqrt{4+h}+2}$.

As $h\to 0$: $\frac{1}{2+2}=\frac{1}{4}$. Final answer: $f^{\prime}(4)=\frac{1}{4}$.

Exercise 7 – Derivative of a cube

$(2+h)^{3}=8+12h+6h^{2}+h^{3}$, so $\frac{(2+h)^{3}-8}{h}=12+6h+h^{2}$.

As $h\to 0$ this approaches $12$. Final answer: $f^{\prime}(2)=12$.

Exercise 8 – The x-approaches-a form

$f(3)=-6$. $\frac{x^{2}-5x+6}{x-3}=\frac{(x-2)(x-3)}{x-3}=x-2$ for $x\neq 3$.

As $x\to 3$: $3-2=1$. Final answer: $f^{\prime}(3)=1$.

Exercise 9 – Derivative function and its zeros

$f(x+h)-f(x)=3(2xh+h^{2})-(3x^{2}h+3xh^{2}+h^{3})=6xh-3x^{2}h+3h^{2}-3xh^{2}-h^{3}$.

Divide by $h$: $6x-3x^{2}+3h-3xh-h^{2}$. As $h\to 0$, $f^{\prime}(x)=6x-3x^{2}=3x(2-x)$.

Final answer: $f^{\prime}(x)=6x-3x^{2};\ \text{horizontal at }x=0\text{ and }x=2$.

Exercise 10 – Using the derivative function

$f(x+h)-f(x)=2(2xh+h^{2})-h=4xh+2h^{2}-h$. Divide by $h$: $4x+2h-1$.

So $f^{\prime}(x)=4x-1$ and $f^{\prime}(-2)=4(-2)-1=-9$. Final answer: $f^{\prime}(x)=4x-1;\ f^{\prime}(-2)=-9$.

Exercises 11 and up

Exercise 11 – Tangent line to a reciprocal

$f(2)=\frac{1}{2}$. $\frac{f(2+h)-f(2)}{h}=\frac{\frac{1}{2+h}-\frac{1}{2}}{h}=\frac{-1}{2(2+h)}\to-\frac{1}{4}$.

Point-slope form: $y-\frac{1}{2}=-\frac{1}{4}(x-2)$.

Final answer: $y=-\frac{1}{4}x+1$.

The tangent to 1/x at x = 2.

Exercise 12 – Tangent line, second example

$f(1)=-2$. $f(1+h)-f(1)=(1+2h+h^{2})-4-4h+1-(-2)=-2h+h^{2}$, so the quotient is $-2+h\to-2$.

Tangent: $y+2=-2(x-1)$, so $y=-2x$. Final answer: $y=-2x$.

The tangent line y = −2x at (1, −2).

Exercise 13 – Slope from a graph

The derivative equals the slope of the tangent: $\frac{3-(-1)}{4-0}=1$.

The tangent line is $y-1=1(x-2)$, which is $y=x-1$.

Final answer: $f^{\prime}(2)=1$ and $y=x-1$.

Exercise 14 – Estimating from a table

$\frac{f(2.1)-f(1.9)}{2.1-1.9}=\frac{7.161-4.959}{0.2}=11.01$.

Final answer: $f^{\prime}(2)\approx 11.01$.

Exercise 15 – A corner

For $h > 0$: $\frac{|h|-0}{h}=1$. For $h < 0$: $\frac{|h|-0}{h}=-1$.

The one-sided limits $1$ and $-1$ differ, so f is not differentiable at x = 1 (even though it is continuous there).

Exercise 16 – A vertical tangent

$\frac{f(0+h)-f(0)}{h}=\frac{h^{1/3}}{h}=h^{-2/3}=\frac{1}{h^{2/3}}$.

As $h\to 0$, $\frac{1}{h^{2/3}}\to\infty$. The tangent line is vertical, so the slope is undefined.

Final answer: not differentiable at x = 0.

Exercise 17 – Continuous but not differentiable

Continuity: left limit $1$, right limit $3-2=1$, $f(1)=1$. So continuous.

Slopes: the left piece has derivative $2x=2$ at $1$; the right piece has slope $3$. They differ.

Final answer: continuous but not differentiable at $x=1$ (a corner).

Exercise 18 – Making a function differentiable

Slopes must match: the derivative of $x^{2}$ at $2$ is $4$, so $a=4$.

Values must match: $2a+b=4$, so $8+b=4$ and $b=-4$.

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

With a = 4 and b = -4 the pieces join smoothly.

Exercise 19 – Speed from the definition

$s(3)=36$ and $s(3+h)=4(9+6h+h^{2})=36+24h+4h^{2}$.

$\frac{24h+4h^{2}}{h}=24+4h\to 24$. Final answer: $24\text{ ft/s}$.

Exercise 20 – Marginal cost

$C(10)=50+200=250$. $C(10+h)=0.5(100+20h+h^{2})+200+20h=250+30h+0.5h^{2}$.

$\frac{30h+0.5h^{2}}{h}=30+0.5h\to 30$.

Final answer: $C^{\prime}(10)=30$: at 10 shirts, each additional shirt costs about \$30.

Exercise 21 – Absolute value at zero

$\frac{|h|-0}{h}$ equals $1$ for $h > 0$ and $-1$ for $h < 0$. The one-sided limits differ, so $f^{\prime}(0)$ does not exist.

For $x > 0$, $f(x)=x$ so $f^{\prime}(x)=1$. For $x < 0$, $f(x)=-x$ so $f^{\prime}(x)=-1$.

Final answer: $f^{\prime}(0)$ does not exist; $f^{\prime}=1$ for $x > 0$ and $f^{\prime}=-1$ for $x < 0$.

Exercise 22 – Recognizing a derivative

This is $f^{\prime}(2)$ for $f(x)=x^{5}$, and $f^{\prime}(x)=5x^{4}$, so the limit is $5\cdot 16=80$.
This is $f^{\prime}(e)$ for $f(x)=\ln x$ since $\ln e=1$. With $f^{\prime}(x)=\frac{1}{x}$, the limit is $\frac{1}{e}$.

Key facts behind this derivative definition answers

Key facts

The derivative is the limit of secant slopes, which are average rates of change.
f′(a) is the slope of the tangent line at x = a; the tangent is y – f(a) = f′(a)(x – a).
Both forms of the definition give the same value when the limit exists.
Differentiable implies continuous, but continuous does not imply differentiable.

How to use this derivative definition 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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