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# 3.7 E: Chain Rule Exercises

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## 3.7: The Chain Rule

#### Exercise:

For the following exercises, given $$y=f(u)$$ and $$u=g(x)$$, find dydx by using Leibniz’s notation for the chain rule: $$\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}.$$

214) $$y=3u−6,u=2x^2$$

215) $$y=6u^3,u=7x−4$$

Solution: $$18u^2⋅7=18(7x−4)^2⋅7$$

216) $$y=sinu,u=5x−1$$

217) $$y=cosu,u=\frac{−x}{8}$$

Solution: $$−sinu⋅\frac{−1}{8=}−sin(\frac{−x}{8})⋅\frac{−1}{8}$$

218) $$y=tanu,u=9x+2$$

219) $$y=\sqrt{4u+3},u=x^2−6x$$

Solution: $$\frac{8x−24}{2\sqrt{4u+3}}=\frac{4x−12}{\sqrt{4x^2−24x+3}}$$

For each of the following exercises,

a. decompose each function in the form $$y=f(u)$$ and $$u=g(x),$$ and

b. find $$\frac{dy}{dx}$$ as a function of $$x$$.

220) $$y=(3x−2)^6$$

221) $$y=(3x^2+1)^3$$

Solution: a. $$u=3x^2+1$$; b. $$18x(3x^2+1)^2$$

222) $$y=sin^5(x)$$

For each of the following exercises, find $$\frac{dy}{dx}$$ as a function of $$x$$.

223) $$y=(\frac{x}{7}+\frac{7}{x})^7$$

Solution: $$a. f(u)=u^7,u=\frac{x}{7}+\frac{7}{x}; b. 7(\frac{x}{7}+\frac{7}{x})^6⋅(\frac{1}{7}−\frac{7}{x^2})$$

224) $$y=tan(secx)$$

225) $$y=csc(πx+1)$$

Solution: $$a. f(u)=cscu,u=πx+1; b. −πcsc(πx+1)⋅cot(πx+1)$$

226) $$y=cot^2x$$

227) $$y=−6sin^{−3}x$$

a. $$f(u)=−6u^{−3},u=sinx, b. 18sin^{−4}x⋅cosx$$

For the following exercises, find $$\frac{dy}{dx}$$ for each function.

228) $$y=(3x^2+3x−1)^4$$

229) $$y=(5−2x)^{−2}$$

Solution: $$\frac{4}{(5−2x)^3}$$

230) $$y=cos^3(πx)$$

231) $$y=(2x^3−x^2+6x+1)^3$$

Solution: $$6(2x^3−x^2+6x+1)^2(3x^2−x+3)$$

232) $$y=\frac{1}{sin^2(x)}$$

233) $$y=(tanx+sinx)^{−3}$$

Soution: $$−3(tanx+sinx)^{−4}⋅(sec^2x+cosx)$$

234) $$y=x^2cos^4x$$

235) $$y=sin(cos7x)$$

Solution: $$−7cos(cos7x)⋅sin7x$$

236) $$y=\sqrt{6+secπx^2}$$

237) $$y=cot^3(4x+1)$$

Solution: $$−12cot^2(4x+1)⋅csc^2(4x+1)$$

238) Let $$y=[f(x)]^3$$ and suppose that $$f′(1)=4$$ and $$\frac{dy}{dx}=10$$ for $$x=1$$. Find $$f(1)$$.

239) Let $$y=(f(x)+5x^2)^4$$ and suppose that $$f(−1)=−4$$ and $$\frac{dy}{dx}=3$$ when $$x=−1$$. Find $$f′(−1)$$

Solution: $$10\frac{3}{4}$$

240) Let $$y=(f(u)+3x)^2$$ and $$u=x^3−2x$$. If $$f(4)=6$$ and $$\frac{dy}{dx}=18$$ when $$x=2$$, find $$f′(4)$$.

241) [T] Find the equation of the tangent line to $$y=−sin(\frac{x}{2})$$ at the origin. Use a calculator to graph the function and the tangent line together.

Solution: $$y=\frac{−1}{2}x$$

242) [T] Find the equation of the tangent line to $$y=(3x+\frac{1}{x})^2$$ at the point $$(1,16)$$. Use a calculator to graph the function and the tangent line together.

243) Find the $$x$$ -coordinates at which the tangent line to $$y=(x−\frac{6}{x})^8$$ is horizontal.

Solution: $$x=±\sqrt{6}$$

244) [T] Find an equation of the line that is normal to $$g(θ)=sin2^(πθ)$$ at the point $$(\frac{1}{4},\frac{1}{2})$$. 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 $$h′(a)$$ at the given value for $$a$$.

 $$x$$ $$f(x)$$ $$f'(x)$$ $$g(x)$$ $$g'(x)$$ 0 2 5 0 2 1 1 −2 3 0 2 4 4 1 −1 3 3 −3 2 3

245) $$h(x)=f(g(x));a=0$$

Solution: $$10$$

246) $$h(x)=g(f(x));a=0$$

247) $$h(x)=(x^4+g(x))^{−2};a=1$$

Solution: $$−\frac{1}{8}$$

248) $$h(x)=(\frac{f(x)}{g(x)})^2;a=3$$

249) $$h(x)=f(x+f(x));a=1$$

Solution: $$−4$$

250) $$h(x)=(1+g(x))^3;a=2$$

251) $$h(x)=g(2+f(x^2));a=1$$

Solution: $$−12$$

252) h(x)=f(g(sinx));a=0

253) [T] The position function of a freight train is given by

$$s(t)=100(t+1)^{−2}$$, with $$s$$ in meters and $$t$$ in seconds. At time $$t=6$$s, find the train’s

a. velocity and

b. acceleration.

c. Using a. and b. is the train speeding up or slowing down?

Solution: $$a. −\frac{200}{343}$$ m/s, b. $$\frac{600}{2401}$$ m/s^2, c. The train is slowing down since velocity and acceleration have opposite signs.

254) [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:

$$s(t)=−3cos(πt+\frac{π}{4}).$$

a. Determine the position of the spring at $$t=1.5$$ s.

b. Find the velocity of the spring at $$t=1.5$$ s.

255) [T] The total cost to produce $$x$$ boxes of Thin Mint Girl Scout cookies is $$C$$ dollars, where $$C=0.0001x^3−0.02x^2+3x+300.$$ In $$t$$ weeks production is estimated to be $$x=1600+100t$$ boxes.

a. Find the marginal cost $$C′(x).$$

b. Use Leibniz’s notation for the chain rule, $$\frac{dC}{dt}=\frac{dC}{dx}⋅\frac{dx}{dt}$$, to find the rate with respect to time $$t$$ that the cost is changing.

c. Use b. to determine how fast costs are increasing when $$t=2$$ weeks. Include units with the answer.

Solution: $$a. C′(x)=0.0003x^2−0.04x+3$$

$$b. dCdt=100⋅(0.0003x^2−0.04x+3)$$ c. Approximately \$90,300 per week

256) [T] The formula for the area of a circle is $$A=πr^2$$, 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.

a. Suppose $$r=2−\frac{100}{(t+7)^2}$$ where $$t$$ is time in seconds. Use the chain rule $$\frac{dA}{dt}=\frac{dA}{dr}⋅\frac{dr}{dt}$$ to find the rate at which the area is expanding.

b. Use a. to find the rate at which the area is expanding at $$t=4$$ s.

257) [T] The formula for the volume of a sphere is $$S=\frac{4}{3}πr^3$$, where $$r$$ (in feet) is the radius of the sphere. Suppose a spherical snowball is melting in the sun.

a. Suppose $$r=\frac{1}{(t+1)^2}−\frac{1}{12}$$ where t is time in minutes. Use the chain rule $$\frac{dS}{dt}=\frac{dS}{dr}⋅\frac{dr}{dt}$$ to find the rate at which the snowball is melting.

b. Use a. to find the rate at which the volume is changing at $$t=1$$ min.

Solution: $$a. \frac{dS}{dt}=−\frac{8πr^2}{(t+1)^3}$$ b. The volume is decreasing at a rate of $$−\frac{π}{36}$$ $$ft^3$$/min

258) [T] The daily temperature in degrees Fahrenheit of Phoenix in the summer can be modeled by the function $$T(x)=94−10cos[\frac{π}{12}(x−2)]$$, where $$x$$ is hours after midnight. Find the rate at which the temperature is changing at 4 p.m.

259) [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 $$D(t)=5sin(\frac{π}{6}t−\frac{7π}{6})+8$$, where $$t$$ is the number of hours after midnight. Find the rate at which the depth is changing at 6 a.m.

Solution: $$~2.3$$ ft/hr

## More Exercises

For the following exercises, find $$f′(x)$$ for each function.

331) $$f(x)=x^2e^x$$

Solution: $$2xe^x+x^2e^x$$

332) $$f(x)=\frac{e^{−x}}{x}$$

333) $$f(x)=e^{x^3lnx}$$

Solution: $$e^{x^3}lnx(3x^2lnx+x^2)$$

334) $$f(x)=\sqrt{e^{2x}+2x}$$

335) $$f(x)=\frac{e^x−e^{−x}}{e^x+e^{−x}}$$

Solution: $$\frac{4}{(e^x+e^{−x})^2}$$

336) $$f(x)=\frac{10^x}{ln10}$$

337) $$f(x)=2^{4x}+4x^2$$

Solution: $$2^{4x+2}⋅ln2+8x$$