Practice questions

Differentiate one term at a time. Use the power rule, and the derivative of a constant is ${0}$:

$\begin{align*}&\dfrac{dy}{dx}\\&=3(4x^{4-1})-5(2x^{2-1})+2(1)-0\\&=12x^3-10x+2\end{align*}$

First, rewrite each term as a power of ${x}$:

${f(x)=4x^{-3}+6x^{1/2}}$

Now differentiate one term at a time:

$\begin{align*}&f'(x)\\&=4(-3x^{-3-1})+6\left(\tfrac{1}{2}x^{1/2-1}\right)\\&=-12x^{-4}+3x^{-1/2}\\&=-\dfrac{12}{x^4}+\dfrac{3}{\sqrt{x}}\end{align*}$

The point: at ${x=1}$, ${y=1-3+4=2}$. So the point is ${(1,2)}$.

The slope: differentiate, then put in ${x=1}$:

$\begin{align*}&y'\\&=3x^{3-1}-3(2x^{2-1})+0\\&=3x^2-6x\end{align*}$

At ${x=1}$, the slope is ${3(1)^2-6(1)=-3}$.

The tangent line:

$\begin{align*}y-2&=-3(x-1)\\y&=-3x+5\end{align*}$

The first factor is ${3x-1}$. Its derivative is ${3}$.

The second factor is ${x^2+4}$. Its derivative is ${2x}$.

Use the product rule:

${y'=3(x^2+4)+(3x-1)(2x)}$

Multiply out each product, and combine like terms:

$\begin{align*}&y'\\&=3x^2+12+6x^2-2x\\&=9x^2-2x+12\end{align*}$

Check: Expand first: ${y=3x^3-x^2+12x-4}$. Then ${y'=9x^2-2x+12}$. The answers match.

The numerator is ${2x+3}$. Its derivative is ${2}$.

The denominator is ${x^2+1}$. Its derivative is ${2x}$.

Use the quotient rule:

$\begin{align*}&y'\\&=\dfrac{2(x^2+1)-(2x+3)(2x)}{(x^2+1)^2}\\&=\dfrac{2x^2+2-4x^2-6x}{(x^2+1)^2}\\&=\dfrac{-2x^2-6x+2}{(x^2+1)^2}\end{align*}$

This is a product. The first factor is ${x^2}$, with derivative ${2x}$. The second factor is ${\sin x}$, with derivative ${\cos x}$. Use the product rule:

${y'=2x\sin x+x^2\cos x}$

Differentiate one term at a time. The derivative of ${\cos x}$ is ${-\sin x}$, and the derivative of ${\sec x}$ is ${\sec x\tan x}$:

$\begin{align*}&y'\\&=3(-\sin x)-2\sec x\tan x\\&=-3\sin x-2\sec x\tan x\end{align*}$

The point: ${\cos\dfrac{\pi}{2}=0}$. So the point is ${\left(\dfrac{\pi}{2},0\right)}$.

The slope: ${y'=-\sin x}$. At ${x=\dfrac{\pi}{2}}$, the slope is ${-\sin\dfrac{\pi}{2}=-1}$.

The tangent line:

$\begin{align*}y-0&=-1\left(x-\dfrac{\pi}{2}\right)\\y&=-x+\dfrac{\pi}{2}\end{align*}$

Let ${u=4x^2-1}$, so ${u'=8x}$.

The outer function is ${u^3}$. Its derivative is ${3u^{3-1}=3u^2}$. Multiply by ${u'}$:

$\begin{align*}&y'\\&=3(4x^2-1)^2\cdot 8x\\&=24x(4x^2-1)^2\end{align*}$

Rewrite the root as a power:

${y=(1+3x^2)^{1/2}}$

Let ${u=1+3x^2}$, so ${u'=6x}$.

$\begin{align*}&y'\\&=\tfrac{1}{2}(1+3x^2)^{1/2-1}\cdot 6x\\&=\tfrac{1}{2}(1+3x^2)^{-1/2}\cdot 6x\\&=\dfrac{3x}{\sqrt{1+3x^2}}\end{align*}$

Let ${u=5x^2}$, so ${u'=10x}$.

The outer function is ${\sin u}$. Its derivative is ${\cos u}$. Multiply by ${u'}$:

$\begin{align*}&y'\\&=\cos(5x^2)\cdot 10x\\&=10x\cos(5x^2)\end{align*}$

First, ${\cos^3 x}$ means ${(\cos x)^3}$.

Let ${u=\cos x}$, so ${u'=-\sin x}$.

The outer function is ${u^3}$. Its derivative is ${3u^2}$. Multiply by ${u'}$:

$\begin{align*}&y'\\&=3\cos^2 x\cdot(-\sin x)\\&=-3\sin x\cos^2 x\end{align*}$

For ${e^{3x}}$, use the chain rule. Let ${u=3x}$, so ${u'=3}$. The derivative of ${e^u}$ is ${e^u}$. Multiply by ${u'}$:

${\dfrac{d}{dx}\left(e^{3x}\right)=3e^{3x}}$

For ${5^x}$, use ${\dfrac{d}{dx}(a^x)=a^x\ln a}$:

${\dfrac{d}{dx}\left(5^x\right)=5^x\ln 5}$

So

${y'=3e^{3x}+5^x\ln 5}$

This is a product. The first factor is ${x^2}$, with derivative ${2x}$. The second factor is ${e^x}$, with derivative ${e^x}$. Use the product rule:

$\begin{align*}&y'\\&=2xe^x+x^2e^x\\&=xe^x(2+x)\end{align*}$

Let ${u=x^2+4}$, so ${u'=2x}$.

The derivative of ${\ln u}$ is ${\dfrac{1}{u}}$. Multiply by ${u'}$:

$\begin{align*}&y'\\&=\dfrac{1}{x^2+4}\cdot 2x\\&=\dfrac{2x}{x^2+4}\end{align*}$

Use $\dfrac{d}{dx}(\log_a x)=\dfrac{1}{x\ln a}$, with ${a=3}$:

${y'=\dfrac{1}{x\ln 3}}$

The variable is in the base and in the exponent. So use logarithmic differentiation.

Take ${\ln}$ of both sides, and use ${\ln(a^p)=p\ln a}$:

${\ln y=2x\ln x}$

Differentiate both sides. The right side needs the product rule:

$\begin{align*}\dfrac{y'}{y}&=2\ln x+2x\cdot\dfrac{1}{x}\\\dfrac{y'}{y}&=2\ln x+2\end{align*}$

Multiply both sides by ${y=x^{2x}}$:

${y'=x^{2x}(2\ln x+2)}$

First, check the point: ${1+2+4=7}$. So ${(1,2)}$ is on the curve.

Differentiate both sides with respect to ${x}$. The term ${xy}$ is a product, so use the product rule:

$\begin{align*}&\dfrac{d}{dx}(xy)\\&=1\cdot y+x\cdot y'\\&=y+xy'\end{align*}$

The derivative of ${x^2}$ is ${2x}$. For ${y^2}$, use the chain rule: its derivative is ${2yy'}$. The right side, ${7}$, is a constant, so its derivative is ${0}$. So the equation becomes

${2x+y+xy'+2yy'=0}$.

Keep the ${y'}$ terms on the left, and move the other terms to the right:

${xy'+2yy'=-2x-y}$

Factor out ${y'}$, and divide:

$\begin{align*}y'(x+2y)&=-2x-y\\y'&=-\dfrac{2x+y}{x+2y}\end{align*}$

At ${(1,2)}$, the slope is

${y'=-\dfrac{2(1)+2}{1+2(2)}=-\dfrac{4}{5}}$

The tangent line:

$\begin{align*}y-2&=-\dfrac{4}{5}(x-1)\\y&=-\dfrac{4}{5}x+\dfrac{14}{5}\end{align*}$

Differentiate both sides with respect to ${x}$. Each ${y}$ term needs the chain rule, which gives a factor ${y'}$:

${3y^2y'+2y'=2x}$

Factor out ${y'}$, and divide:

$\begin{align*}y'(3y^2+2)&=2x\\y'&=\dfrac{2x}{3y^2+2}\end{align*}$

Differentiate three times, one step at a time:

$\begin{align*}&f'(x)\\&=4x^{4-1}-2(3x^{3-1})+5\\&=4x^3-6x^2+5\end{align*}$

$\begin{align*}&f''(x)\\&=4(3x^{3-1})-6(2x^{2-1})+0\\&=12x^2-12x\end{align*}$

$\begin{align*}&f'''(x)\\&=12(2x^{2-1})-12\\&=24x-12\end{align*}$

Use the chain rule with ${u=2x}$, so ${u'=2}$:

${y'=\cos 2x\cdot 2=2\cos 2x}$

Differentiate again. The derivative of ${\cos u}$ is ${-\sin u}$:

$\begin{align*}&y''\\&=2(-\sin 2x)\cdot 2\\&=-4\sin 2x\end{align*}$

The velocity is the derivative of the position. The acceleration is the derivative of the velocity:

$\begin{align*}v(t)&=3t^2-18t+24\\a(t)&=6t-18\end{align*}$

The particle is at rest when its velocity is ${0}$:

$\begin{align*}3t^2-18t+24&=0\\t^2-6t+8&=0\\(t-2)(t-4)&=0\end{align*}$

So it is at rest at ${t=2}$ s and at ${t=4}$ s.

The acceleration at those times:

${a(2)=12-18=-6}$ m/s${^2}$

${a(4)=24-18=6}$ m/s${^2}$

At ${t=2}$, the particle stops and starts moving backward. At ${t=4}$, it stops again and starts moving forward.