3.6E: Exercises for Section 3.6

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In exercises 1 - 6, given $y=f(u)$ and $u=g(x)$, find $\dfrac{dy}{dx}$ by using Leibniz’s notation for the chain rule: $\dfrac{dy}{dx}=\dfrac{dy}{du}\dfrac{du}{dx}.$

1) $y=3u−6,\quad u=2x^2$

2) $y=6u^3,\quad u=7x−4$

Answer: $\dfrac{dy}{dx} = 18u^2⋅7=18(7x−4)^2⋅7= 126(7x−4)^2$

3) $y=\sin u,\quad u=5x−1$

4) $y=\cos u,\quad u=-\frac{x}{8}$

Answer: $\dfrac{dy}{dx} = −\sin u⋅\left(-\frac{1}{8}\right)=\frac{1}{8}\sin(-\frac{x}{8})$

5) $y=\tan u,\quad u=9x+2$

6) $y=\sqrt{4u+3},\quad u=x^2−6x$

Answer: $\dfrac{dy}{dx} = \dfrac{8x−24}{2\sqrt{4u+3}}=\dfrac{4x−12}{\sqrt{4x^2−24x+3}}$

For each of the following exercises,

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

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

7) $y=(3x−2)^6$

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

Answer: a. $f(u)=u^3,\quad u=3x^2+1$;

b. $\dfrac{dy}{dx} = 18x(3x^2+1)^2$

9) $y=\sin^5(x)$

10) $y=\left(\dfrac{x}{7}+\dfrac{7}{x}\right)^7$

Answer: a. $f(u)=u^7,\quad u=\dfrac{x}{7}+\dfrac{7}{x}$;

b. $\dfrac{dy}{dx} = 7\left(\dfrac{x}{7}+\dfrac{7}{x}\right)^6⋅\left(\dfrac{1}{7}−\dfrac{7}{x^2}\right)$

11) $y=\tan(\sec x)$

12) $y=\csc(πx+1)$

Answer: a. $f(u)=\csc u,\quad u=πx+1$;

b. $\dfrac{dy}{dx} = −π\csc(πx+1)⋅\cot(πx+1)$

13) $y=\cot^2x$

14) $y=−6\sin^{−3}x$

Answer: a. $f(u)=−6u^{−3},\quad u=\sin x$;

b. $\dfrac{dy}{dx} = 18\sin^{−4}x⋅\cos x$

In exercises 15 - 24, find $\dfrac{dy}{dx}$ for each function.

15) $y=(3x^2+3x−1)^4$

16) $y=(5−2x)^{−2}$

Answer: $\dfrac{dy}{dx}=\dfrac{4}{(5−2x)^3}$

17) $y=\cos^3(πx)$

18) $y=(2x^3−x^2+6x+1)^3$

Answer: $\dfrac{dy}{dx}=6(2x^3−x^2+6x+1)^2⋅(3x^2−x+3)$

19) $y=\dfrac{1}{\sin^2(x)}$

20) $y=\big(\tan x+\sin x\big)^{−3}$

Answer: $\dfrac{dy}{dx}=−3\big(\tan x+\sin x\big)^{−4}⋅(\sec^2x+\cos x)$

21) $y=x^2\cos^4x$

22) $y=\sin(\cos 7x)$

Answer: $\dfrac{dy}{dx}=−7\cos(\cos 7x)⋅\sin 7x$

23) $y=\sqrt{6+\sec πx^2}$

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

Answer: $\dfrac{dy}{dx}=−12\cot^2(4x+1)⋅\csc^2(4x+1)$

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

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

Answer: $f′(−1)=10\frac{3}{4}$

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

28) [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.

Answer: $y=-\frac{1}{2}x$

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

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

Answer: $x=±\sqrt{6}$

31) [T] Find an equation of the line that is normal to $g(θ)=\sin^2(πθ)$ at the point $\left(\frac{1}{4},\frac{1}{2}\right)$. Use a calculator to graph the function and the normal line together.

For exercises 32 - 39, 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

32) $h(x)=f\big(g(x)\big);\quad a=0$

Answer: $h'(0) = 10$

33) $h(x)=g\big(f(x)\big);\quad a=0$

34) $h(x)=\big(x^4+g(x)\big)^{−2};\quad a=1$

Answer: $h'(1) = −\frac{1}{8}$

35) $h(x)=\left(\dfrac{f(x)}{g(x)}\right)^2;\quad a=3$

36) $h(x)=f\big(x+f(x)\big);\quad a=1$

Answer: $h'(1) = −4$

37) $h(x)=\big(1+g(x)\big)^3;\quad a=2$

38) $h(x)=g\big(2+f(x^2)\big);\quad a=1$

Answer: $h'(1) = −12$

39) $h(x)=f\big(g(\sin x)\big);\quad a=0$

40) [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. Considering your results in parts a. and b., is the train speeding up or slowing down?

Answer: a. $v(6) = −\frac{200}{343}$ m/s,

b. $a(6) = \frac{600}{2401}\;\text{m/s}^2,$

c. The train is slowing down since velocity and acceleration have opposite signs.

41) [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:

\[s(t)=−3\cos\left(πt+\frac{π}{4}\right).\nonumber \]

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

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

42) [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, $\dfrac{dC}{dt}=\dfrac{dC}{dx}⋅\dfrac{dx}{dt}$, to find the rate with respect to time $t$ that the cost is changing.

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

Answer: a. $C′(x)=0.0003x^2−0.04x+3$

b. $\dfrac{dC}{dt}=100⋅(0.0003x^2−0.04x+3) = 100⋅(0.0003(1600+100t)^2−0.04(1600+100t)+3) = 300t^2 +9200t +70700$

c. Approximately $90,300 per week

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

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

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

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

Answer: a. $\dfrac{dS}{dt}=−\dfrac{8πr^2}{(t+1)^3} = −\dfrac{8π\left( \dfrac{1}{(t+1)^2}−\dfrac{1}{12} \right)^2}{(t+1)^3}$

b. The volume is decreasing at a rate of $−\frac{π}{36}\; \text{ft}^3$/min

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

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

Answer: $~2.3$ ft/hr

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

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