2.4.3: Additional Exercises
- Page ID
- 116569
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- State the Chain Rule using Lagrange's notation for a composite function \(h(x) = f(g(x))\).
- State the Chain Rule using Leibniz's notation if \(y\) is a function of \(u\) and \(u\) is a function of \(x\).
- What is the General Power Rule (Power Rule for Composition of Functions) for differentiating \(h(x) = (g(x))^n\)?
- If \(h(x) = \sin(x^3)\), identify the "outer" function \(f(u)\) and the "inner" function \(u=g(x)\). Then find \(h^{\prime}(x)\).
- How is the Chain Rule applied to find the derivative of a composition of three functions, say \(k(x) = h(f(g(x)))\)?
- When differentiating a function like \(h(x) = \frac{1}{(3x^2+1)^2}\), what algebraic step is recommended before applying the General Power Rule?
- In Example 7, \(h(x) = (2x+1)^5(3x-2)^7\), which differentiation rule is applied first, and which rule is then applied to the terms resulting from the first rule?
- When using Leibniz notation for the Chain Rule, like \(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\), what must be done to the term \(\frac{dy}{du}\) to express the final answer in terms of \(x\)?
Homework
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
- \(\dfrac{dy}{dx} = 18u^2 \cdot 7=18(7x−4)^2 \cdot 7= 126(7x−4)^2\)
3) \(y=\sin u,\quad u=5x−1\)
4) \(y=\cos u,\quad u=-\frac{x}{8}\)
- Answer
- \(\dfrac{dy}{dx} = −\sin u \cdot \left(-\frac{1}{8}\right)=\frac{1}{8}\sin(-\frac{x}{8})\)
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. \(\dfrac{dy}{dx} = 7\left(\dfrac{x}{7}+\dfrac{7}{x}\right)^6 \cdot \left(\dfrac{1}{7}−\dfrac{7}{x^2}\right)\)
11) \(y=\tan(\sec x)\)
12) \(y=\csc( \pi x+1)\)
- Answer
- a. \(f(u)=\csc u,\quad u= \pi x+1\);
b. \(\dfrac{dy}{dx} = − \pi \csc( \pi x+1) \cdot \cot( \pi x+1)\)
13) \(y=\cot^2x\)
14) \(y=−6\sin^{−3}x\)
- Answer
- a. \(f(u)=−6u^{−3},\quad u=\sin x\);
b. \(\dfrac{dy}{dx} = 18\sin^{−4}x \cdot \cos x\)
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) \(y=\cos^3( \pi x)\)
18) \(y=(2x^3−x^2+6x+1)^3\)
- Answer
- \(\dfrac{dy}{dx}=6(2x^3−x^2+6x+1)^2 \cdot (3x^2−x+3)\)
19) \(y=\dfrac{1}{\sin^2(x)}\)
20) \(y=\big(\tan x+\sin x\big)^{−3}\)
- Answer
- \(\dfrac{dy}{dx}=−3\big(\tan x+\sin x\big)^{−4} \cdot (\sec^2x+\cos x)\)
21) \(y=x^2\cos^4x\)
22) \(y=\sin(\cos 7x)\)
- Answer
- \(\dfrac{dy}{dx}=−7\cos(\cos 7x) \cdot \sin 7x\)
23) \(y=\sqrt{6+\sec \pi x^2}\)
24) \(y=\cot^3(4x+1)\)
- Answer
- \(\dfrac{dy}{dx}=−12\cot^2(4x+1) \cdot \csc^2(4x+1)\)
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) [Technology Required] 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) [Technology Required] 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
- \(x= \pm \sqrt{6}\)
31) [Technology Required] Find an equation of the line that is normal to \(g( \theta )=\sin^2( \pi \theta )\) 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 \(a\).
| \(x\) | \(f(x)\) | \(f^{\prime}(x)\) | \(g(x)\) | \(g^{\prime}(x)\) |
| 0 | 2 | 5 | 0 | 2 |
| 1 | 1 | −2 | 3 | 0 |
| 2 | 4 | 4 | 1 | −1 |
| 3 | 3 | −3 | 2 | 3 |
32) \(h(x)=f\big(g(x)\big);\quad a=0\)
- Answer
- \(h^{\prime}(0) = 10\)
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
- \(h^{\prime}(1) = −\frac{1}{8}\)
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
- \(h^{\prime}(1) = −4\)
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
- \(h^{\prime}(1) = −12\)
39) \(h(x)=f\big(g(\sin x)\big);\quad a=0\)
40) [Technology Required] 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) [Technology Required] 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:
\[s(t)=−3\cos\left( \pi t+\frac{ \pi }{4}\right).\nonumber \]
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) [Technology Required] 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, \(\dfrac{dC}{dt}=\dfrac{dC}{dx} \cdot \dfrac{dx}{dt}\), 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. \(\dfrac{dC}{dt}=100 \cdot (0.0003x^2−0.04x+3) = 100 \cdot (0.0003(1600+100t)^2−0.04(1600+100t)+3) = 300t^2 +9200t +70700\)
c. Approximately $90,300 per week
43) [Technology Required] 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.
a. Suppose \(r=2−\dfrac{100}{(t+7)^2}\) where \(t\) is time in seconds. Use the chain rule \(\dfrac{dA}{dt}=\dfrac{dA}{dr} \cdot \dfrac{dr}{dt}\) 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) [Technology Required] 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. Suppose \(r=\dfrac{1}{(t+1)^2}−\dfrac{1}{12}\) where \(t\) is time in minutes. Use the chain rule \(\dfrac{dS}{dt}=\dfrac{dS}{dr} \cdot \dfrac{dr}{dt}\) 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. \(\dfrac{dS}{dt}=−\dfrac{8 \pi r^2}{(t+1)^3} = −\dfrac{8 \pi \left( \dfrac{1}{(t+1)^2}−\dfrac{1}{12} \right)^2}{(t+1)^3}\)
b. The volume is decreasing at a rate of \(−\frac{ \pi }{36}\; \text{ft}^3\)/min
45) [Technology Required] The daily temperature in degrees Fahrenheit of Phoenix in the summer can be modeled by the function \(T(x)=94−10\cos\left[\frac{ \pi }{12}(x−2)\right]\), where \(x\) is hours after midnight. Find the rate at which the temperature is changing at 4 p.m.
46) [Technology Required] The depth (in feet) of water at a dock changes with the rise and fall of tides. The depth is modeled by the function \(D(t)=5\sin\left(\frac{ \pi }{6}t−\frac{7 \pi }{6}\right)+8\), where \(t\) is the number of hours after midnight. Find the rate at which the depth is changing at 6 a.m.
- Answer
- \(~2.3\) ft/hr