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Find the derivative of $y={\left(\frac{x}{3x+2}\right)}^{5}.$
First, let $u=\frac{x}{3x+2}.$ Thus, $y={u}^{5}.$ Next, find $\frac{du}{dx}$ and $\frac{dy}{du}.$ Using the quotient rule,
and
Finally, we put it all together.
It is important to remember that, when using the Leibniz form of the chain rule, the final answer must be expressed entirely in terms of the original variable given in the problem.
Find the derivative of $y=\text{tan}\phantom{\rule{0.1em}{0ex}}\left(4{x}^{2}-3x+1\right).$
First, let $u=4{x}^{2}-3x+1.$ Then $y=\text{tan}\phantom{\rule{0.1em}{0ex}}u.$ Next, find $\frac{du}{dx}$ and $\frac{dy}{du}\text{:}$
Finally, we put it all together.
Use Leibniz’s notation to find the derivative of $y=\text{cos}\phantom{\rule{0.1em}{0ex}}\left({x}^{3}\right).$ Make sure that the final answer is expressed entirely in terms of the variable $x.$
$\frac{dy}{dx}=\mathrm{-3}{x}^{2}\text{sin}\phantom{\rule{0.1em}{0ex}}\left({x}^{3}\right)$
For the following exercises, given $y=f(u)$ and $u=g\left(x\right),$ find $\frac{dy}{dx}$ by using Leibniz’s notation for the chain rule: $\frac{dy}{dx}=\frac{dy}{du}\phantom{\rule{0.2em}{0ex}}\frac{du}{dx}.$
$y=3u-6,u=2{x}^{2}$
$y=6{u}^{3},u=7x-4$
$18{u}^{2}\xb77=18{\left(7x-4\right)}^{2}\xb77$
$y=\text{sin}\phantom{\rule{0.1em}{0ex}}u,u=5x-1$
$y=\text{cos}\phantom{\rule{0.1em}{0ex}}u,u=\frac{\text{\u2212}x}{8}$
$\text{\u2212}\text{sin}\phantom{\rule{0.1em}{0ex}}u\xb7\frac{\mathrm{-1}}{8}=\text{\u2212}\text{sin}\phantom{\rule{0.1em}{0ex}}\left(\frac{\text{\u2212}x}{8}\right)\xb7\frac{\mathrm{-1}}{8}$
$y=\text{tan}\phantom{\rule{0.1em}{0ex}}u,u=9x+2$
$y=\sqrt{4u+3},u={x}^{2}-6x$
$\frac{8x-24}{2\sqrt{4u+3}}=\frac{4x-12}{\sqrt{4{x}^{2}-24x+3}}$
For each of the following exercises,
$y={\left(3x-2\right)}^{6}$
$y={\left(3{x}^{2}+1\right)}^{3}$
a. $u=3{x}^{2}+1;$ b. $18x{\left(3{x}^{2}+1\right)}^{2}$
$y={\text{sin}}^{5}\left(x\right)$
$y={\left(\frac{x}{7}+\frac{7}{x}\right)}^{7}$
a. $f\left(u\right)={u}^{7},u=\frac{x}{7}+\frac{7}{x};$ b. $7{\left(\frac{x}{7}+\frac{7}{x}\right)}^{6}\xb7\left(\frac{1}{7}-\frac{7}{{x}^{2}}\right)$
$y=\text{tan}\phantom{\rule{0.1em}{0ex}}\left(\text{sec}\phantom{\rule{0.1em}{0ex}}x\right)$
$y=\text{csc}\phantom{\rule{0.1em}{0ex}}\left(\pi x+1\right)$
a. $f\left(u\right)=\text{csc}\phantom{\rule{0.1em}{0ex}}u,u=\pi x+1;$ b. $\text{\u2212}\pi \phantom{\rule{0.1em}{0ex}}\text{csc}\phantom{\rule{0.1em}{0ex}}\left(\pi x+1\right)\xb7\text{cot}\phantom{\rule{0.1em}{0ex}}\left(\pi x+1\right)$
$y={\text{cot}}^{2}x$
$y=\mathrm{-6}\phantom{\rule{0.1em}{0ex}}{\text{sin}}^{\mathrm{-3}}x$
a. $f\left(u\right)=\mathrm{-6}{u}^{\mathrm{-3}},u=\text{sin}\phantom{\rule{0.1em}{0ex}}x,$ b. $18\phantom{\rule{0.1em}{0ex}}{\text{sin}}^{\mathrm{-4}}x\xb7\text{cos}\phantom{\rule{0.1em}{0ex}}x$
For the following exercises, find $\frac{dy}{dx}$ for each function.
$y={\left(3{x}^{2}+3x-1\right)}^{4}$
$y={\left(5-2x\right)}^{\mathrm{-2}}$
$\frac{4}{{\left(5-2x\right)}^{3}}$
$y={\text{cos}}^{3}\left(\pi x\right)$
$y={\left(2{x}^{3}-{x}^{2}+6x+1\right)}^{3}$
$6{\left(2{x}^{3}-{x}^{2}+6x+1\right)}^{2}\left(3{x}^{2}-x+3\right)$
$y=\frac{1}{{\text{sin}}^{2}(x)}$
$y={\left(\text{tan}\phantom{\rule{0.1em}{0ex}}x+\text{sin}\phantom{\rule{0.1em}{0ex}}x\right)}^{\mathrm{-3}}$
$\mathrm{-3}{\left(\text{tan}\phantom{\rule{0.1em}{0ex}}x+\text{sin}\phantom{\rule{0.1em}{0ex}}x\right)}^{\mathrm{-4}}\xb7\left({\text{sec}}^{2}x+\text{cos}\phantom{\rule{0.1em}{0ex}}x\right)$
$y={x}^{2}{\text{cos}}^{4}x$
$y=\text{sin}\phantom{\rule{0.1em}{0ex}}\left(\text{cos}\phantom{\rule{0.1em}{0ex}}7x\right)$
$\mathrm{-7}\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}\left(\text{cos}\phantom{\rule{0.1em}{0ex}}7x\right)\xb7\text{sin}\phantom{\rule{0.1em}{0ex}}7x$
$y=\sqrt{6+\text{sec}\phantom{\rule{0.1em}{0ex}}\pi {x}^{2}}$
$y={\text{cot}}^{3}\left(4x+1\right)$
$\mathrm{-12}\phantom{\rule{0.1em}{0ex}}{\text{cot}}^{2}(4x+1)\xb7{\text{csc}}^{2}(4x+1)$
Let $y={\left[f(x)\right]}^{3}$ and suppose that ${f}^{\prime}\left(1\right)=4$ and $\frac{dy}{dx}=10$ for $x=1.$ Find $f\left(1\right).$
Let $y={\left(f\left(x\right)+5{x}^{2}\right)}^{4}$ and suppose that $f\left(\mathrm{-1}\right)=\mathrm{-4}$ and $\frac{dy}{dx}=3$ when $x=\mathrm{-1}.$ Find ${f}^{\prime}\left(\mathrm{-1}\right)$
$10\frac{3}{4}$
Let $y={\left(f\left(u\right)+3x\right)}^{2}$ and $u={x}^{3}-2x.$ If $f\left(4\right)=6$ and $\frac{dy}{dx}=18$ when $x=2,$ find ${f}^{\prime}\left(4\right).$
[T] Find the equation of the tangent line to $y=\text{\u2212}\text{sin}\phantom{\rule{0.1em}{0ex}}\left(\frac{x}{2}\right)$ at the origin. Use a calculator to graph the function and the tangent line together.
$y=\frac{\mathrm{-1}}{2}x$
[T] Find the equation of the tangent line to $y={\left(3x+\frac{1}{x}\right)}^{2}$ at the point $\left(1,16\right).$ Use a calculator to graph the function and the tangent line together.
Find the $x$ -coordinates at which the tangent line to $y={\left(x-\frac{6}{x}\right)}^{8}$ is horizontal.
$x=\pm \sqrt{6}$
[T] Find an equation of the line that is normal to $g(\theta )={\text{sin}}^{2}\left(\pi \theta \right)$ at the point $\left(\frac{1}{4},\frac{1}{2}\right).$ Use a calculator to graph the function and the normal line together.
For the following exercises, use the information in the following table to find ${h}^{\prime}(a)$ at the given value for $a.$
$x$ | $f(x)$ | $f\prime (x)$ | $g(x)$ | $g\prime (x)$ |
---|---|---|---|---|
0 | 2 | 5 | 0 | 2 |
1 | 1 | −2 | 3 | 0 |
2 | 4 | 4 | 1 | −1 |
3 | 3 | −3 | 2 | 3 |
$h\left(x\right)=g\left(f\left(x\right)\right);a=0$
$h\left(x\right)={\left({x}^{4}+g(x)\right)}^{\mathrm{-2}};a=1$
$-\frac{1}{8}$
$h(x)={\left(\frac{f(x)}{g(x)}\right)}^{2};a=3$
$h\left(x\right)=f\left(x+f\left(x\right)\right);a=1$
$\mathrm{-4}$
$h\left(x\right)={\left(1+g(x)\right)}^{3};a=2$
$h\left(x\right)=g\left(2+f\left({x}^{2}\right)\right);a=1$
$\mathrm{-12}$
$h\left(x\right)=f\left(g\left(\text{sin}\phantom{\rule{0.1em}{0ex}}x\right)\right);a=0$
[T] The position function of a freight train is given by $s\left(t\right)=100{\left(t+1\right)}^{\mathrm{-2}},$ with $s$ in meters and $t$ in seconds. At time $t=6$ s, find the train’s
a. $-\frac{200}{343}$ m/s, b. $\frac{600}{2401}$ m/s ^{2} , c. The train is slowing down since velocity and acceleration have opposite signs.
[T] A mass hanging from a vertical spring is in simple harmonic motion as given by the following position function, where $t$ is measured in seconds and $s$ is in inches:
$s\left(t\right)=\mathrm{-3}\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}\left(\pi t+\frac{\pi}{4}\right).$
[T] The total cost to produce $x$ boxes of Thin Mint Girl Scout cookies is $C$ dollars, where $C=0.0001{x}^{3}-0.02{x}^{2}+3x+300.$ In $t$ weeks production is estimated to be $x=1600+100t$ boxes.
a. ${C}^{\prime}\left(x\right)=0.0003{x}^{2}-0.04x+3$ b. $\frac{dC}{dt}=100\xb7\left(0.0003{x}^{2}-0.04x+3\right)$ c. Approximately $90,300 per week
[T] The formula for the area of a circle is $A=\pi {r}^{2},$ where $r$ is the radius of the circle. Suppose a circle is expanding, meaning that both the area $A$ and the radius $r$ (in inches) are expanding.
[T] The formula for the volume of a sphere is $S=\frac{4}{3}\pi {r}^{3},$ where $r$ (in feet) is the radius of the sphere. Suppose a spherical snowball is melting in the sun.
a. $\frac{dS}{dt}=-\frac{8\pi {r}^{2}}{{\left(t+1\right)}^{3}}$ b. The volume is decreasing at a rate of $-\frac{\pi}{36}$ ft ^{3} /min.
[T] The daily temperature in degrees Fahrenheit of Phoenix in the summer can be modeled by the function $T\left(x\right)=94-10\phantom{\rule{0.1em}{0ex}}\text{cos}\left[\frac{\pi}{12}\left(x-2\right)\right],$ where $x$ is hours after midnight. Find the rate at which the temperature is changing at 4 p.m.
[T] The depth (in feet) of water at a dock changes with the rise and fall of tides. The depth is modeled by the function $D\left(t\right)=5\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}\left(\frac{\pi}{6}t-\frac{7\pi}{6}\right)+8,$ where $t$ is the number of hours after midnight. Find the rate at which the depth is changing at 6 a.m.
$~2.3$ ft/hr
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