AP Calculus AB Unit 3: Differentiation Rules

Explore AP Calculus Unit 3 questions on the chain rule, implicit differentiation, tangent lines, and related rates in context.

Covered syllabus points

[Maximum number: 2]

Two particles, H and J, are moving along the x-axis. For $0 \leq t \leq 5$ , the position of particle H at

time t is given by $x_{H}(t)=e^{t^{2}-4 t}$ and the velocity of particle J at time t is given by

$v_{J}(t)=2 t\left(t^{2}-1\right)^{3}$ .

Find the velocity of particle H at time t=1. Show the work that leads to your answer.

A	Find the velocity of particle H at time t=1. Show the work that leads to your answer.
		$x_{H}{ }^{\prime}(t)=v_{H}(t)=(2 t-4) e^{t^{2}-4 t}$	Considers $x_{H}{ }^{\prime}$ Point 1 (P1)
		$x_{H}{ }^{\prime}(1)=v_{H}(1)=-2 e^{-3}$	Answer Point 2 (P2)
		Scoring Notes for Part A
		- P1 can be earned by presenting $x_{H}{ }^{\prime}(t), x_{H}{ }^{\prime}(1), x^{\prime}(t), x^{\prime}(1),(2 t-4) e^{t^{2}-4 t}$ , or $(2 \cdot 1-4) e^{1^{2}-4 \cdot 1}$ . - An unsupported answer of $-2 e^{-3}$ earns P2 but not P1.

[Maximum number: 2]

Consider the curve G defined by the equation $y^{3}-y^{2}-y+\frac{1}{4} x^{2}=0$ .

Show that $\frac{d y}{d x}=\frac{-x}{2\left(3 y^{2}-2 y-1\right)}$ .


$\begin{aligned} & \frac{d}{d x}\left(y^{3}-y^{2}-y+\frac{1}{4} x^{2}\right)=0 \\ & \Rightarrow 3 y^{2} \frac{d y}{d x}-2 y \frac{d y}{d x}-\frac{d y}{d x}+\frac{x}{2}=0 \end{aligned}$	Implicit differentiation	Point 1 (P1)
$\begin{aligned} & \Rightarrow\left(3 y^{2}-2 y-1\right) \frac{d y}{d x}=-\frac{x}{2} \\ & \Rightarrow \frac{d y}{d x}=\frac{-x}{2\left(3 y^{2}-2 y-1\right)} \end{aligned}$	Verification	Point 2 (P2)
Scoring Notes for Part A
- P1 is earned only for the correct implicit differentiation of $y^{3}-y^{2}-y+\frac{1}{4}x^{2}=0$ . Alternative notation such as $y^{\prime}$ is acceptable. - It is sufficient to present $\left(3 y^{2}-2 y-1\right)\frac{d y}{d x}=-\frac{x}{2}$ to earn P2, provided there are no subsequent errors.

[Maximum number: 1]

Questions 9 and 10. Use the following table, which shows the values of differentiable functions $\boldsymbol{f}$ and $\boldsymbol{g}$ .

If $H(x)=f^{-1}(x)$ , then $H^{\prime}(3)$ equals

$-\frac{1}{16}$

$-\frac{1}{8}$

$\frac{1}{2}$

[Maximum number: 1]

$\frac{d}{d x}\left(\tan ^{-1} x+2 \sqrt{x}\right)=$

$-\frac{1}{\sin ^{2} x}+\frac{1}{\sqrt{x}}$

$\frac{1}{\sqrt{1-x^{2}}}-4 \sqrt[3]{x}$

$\frac{1}{\sqrt{1-x^{2}}}+\frac{1}{\sqrt{x}}$

$\frac{1}{1+x^{2}}-4 \sqrt[3]{x}$

$\frac{1}{1+x^{2}}+\frac{1}{\sqrt{x}}$

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