3.6E: Exercises for Section 3.6

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