Practice Problems

In problems 1-5, find two functions $\mathrm{f}$ and $\mathrm{g}$ so that the given function is the composition of $\mathrm{f}$ and $\mathrm{g}$.

1. $y=\left(x^{3}-7 x\right)^{5}$

3. $y=\sqrt{(2+\sin (x))^{5}}$

5. $y=\left|x^{2}-4\right|$

7. For each function in problems 1-5, write $y$ as a function of $u$ for some $u$ which is a function of $x$. Problems 8 and 9 refer to the values given in this table:

$\begin{array}{|c|c|c|c|c|c|c} \hline x & f(x) & g(x) & f^{\prime}(x) & g^{\prime}(x) & (f \circ g)(x) & \text { ( fog })^{\prime}(x) \\ \hline-2 & 2 & -1 & 1 & 1 & & \\ -1 & 1 & 2 & 0 & 2 & & \\ 0 & -2 & 1 & 2 & -1 & & \\ 1 & 0 & -2 & -1 & 2 & & \\ 2 & 1 & 0 & 1 & -1 & & \end{array}$

9. Use the table of values to determine $(\mathrm{f \circ g})(\mathrm{x})$ and $(\mathrm{f ∘ g})^{\prime}(\mathrm{x})$ at $\mathrm{x}=-2,-1$ and $0$.

11. Use Fig. 2 to estimate the values of $g(x), g^{\prime}(x)$ $(f \circ g)(x), f^{\prime}(g(x))$, and $(f \circ g)^{\prime}(x)$ for $x=2$

In problems 13-19, differentiate each function.

13. $\mathrm{D}\left(\left(1-\frac{3}{\mathrm{x}}\right)^{4}\right)$

15. $\quad \mathbf{D}\left(\frac{5}{\sqrt{2+\sin (x)}}\right)$

17. $\quad D\left(x^{2} \cdot \sin \left(x^{2}+3\right)\right)$

19. $\operatorname{D}\left(\frac{7}{\cos \left(x^{3}-x\right)}\right)$

21. $\quad \mathbf{D}\left(\mathrm{e}^{\mathrm{x}}+\mathrm{e}^{-\mathrm{x}}\right)$

23. An object attached to a spring is at a height of $\mathrm{h}(\mathrm{t})=3-\cos (2 \mathrm{t})$ feet above the floor $\mathrm{t}$ seconds after it is released.

(a) At what height was it released?

(b) Determine its height, velocity and acceleration at any time $t$.

(c) If the object has mass $\mathrm{m}$, determine its kinetic energy $\mathrm{K}=\frac{1}{2} \mathrm{mv}^{2}$ and $\mathrm{dK} / \mathrm{dt}$ at any time $\mathrm{t}$.

25. The air pressure $\mathrm{P}(\mathrm{h})$, in pounds per square inch, at an altitude of $\mathrm{h}$ feet above sea level is approximately $\mathrm{P}(\mathrm{h})=14.7 \mathrm{e}^{-0.0000385 \mathrm{~h}}$

(a) What is the air pressure at sea level? What is the air pressure at an altitude of 30,000 feet?

(b) At what altitude is the air pressure 10 pounds per square inch?

(c) If you are in a balloon which is 2000 feet above the Pacific Ocean and is rising at 500 feet per minute, how fast is the air pressure on the balloon changing?

(d) If the temperature of the gas in the balloon remained constant during this ascent, what would happen to the volume of the balloon?

Find the derivatives in problems 27-33.

27. $\left.\frac{\mathbf{d}}{\mathbf{d z}} \sqrt{1+\cos ^{2}(z)}\right)$

29. $\frac{\mathrm{d}}{\mathrm{dx}} \tan (3 \mathrm{x}+5)$

31. $\mathrm{D}(\sin (\sqrt{\mathrm{x}+1}))$

33. $\frac{\mathbf{d}}{\mathbf{d x}}\left(\mathrm{e}^{\sin (\mathrm{x})}\right)$

In problems 35-37, calculate $\frac{\mathrm{d} \mathrm{f}(\mathrm{x})}{\mathrm{dx}}$ and $\frac{\mathrm{d} \mathrm{x}(\mathrm{t})}{\mathrm{dt}}$ when $\mathrm{t}=3$ and use these values to determine the value of $\frac{\mathrm{d} \mathrm{f}(\mathrm{x}(\mathrm{t}))}{\mathrm{dt}}$ when $\mathrm{t}=3$

35. $\mathrm{f}(\mathrm{x})=\sqrt{\mathrm{x}} \quad, \mathrm{x}=2+\frac{21}{\mathrm{t}}$

37. $\mathrm{f}(\mathrm{x})=\tan ^{3}(\mathrm{x}), \mathrm{x}=8$

In problems 39-43, find a function which has the given function as its derivative. (You are given $\mathrm{f}^{\prime}(\mathrm{x})$ in each problem and are asked to find a function $\mathrm{f}(\mathrm{x})$.)

39. $\mathrm{f}^{\prime}(\mathrm{x})=(7 \mathrm{x}-13)^{10}$

41. $\mathrm{f}^{\prime}(\mathrm{x})=\sin (2 \mathrm{x}-3)$

43. $\mathrm{f}^{\prime}(\mathrm{x})=\cos (\mathrm{x}) \cdot \mathrm{e}^{\sin (\mathrm{x})}$

If two functions are equal, then their derivatives are also equal. In problems 45-47, differentiate each side of the trigonometric identity to find a new identity.

45. $\cos (2 x)=\cos ^{2}(x)-\sin ^{2}(x)$

47. $\sin (3 x)=3 \sin (x)-4 \sin ^{3}(x)$

Source: Dale Hoffman, https://s3.amazonaws.com/saylordotorg-resources/wwwresources/site/wp-content/uploads/2012/12/MA005-3.5-Chain-Rule.pdf
This work is licensed under a Creative Commons Attribution 3.0 License.

So far we have emphasized derivatives of particular functions, but sometimes we want to look at the  derivatives of a whole family of functions.  In problems  49 –71, the letters A – D represent constants and the given formulas describe families of functions. 

For problems 49-65, calculate $\mathrm{y}^{\prime}=\frac{\mathrm{d} \mathrm{y}}{\mathrm{d} \mathrm{x}}$.

49. $y=A x^{3}+B x^{2}+C$

51. $y=\sin \left(A x^{2}+B\right)$

53. $\mathrm{y}=\sqrt{\mathrm{A}+\mathrm{Bx}^{2}}$

55. $y=A-\cos (B x)$

57. $y=\cos \left(A x^{2}+B\right)$

59. $y=x \cdot e^{B x}$

61. $y=e^{A x}-e^{-A x}$

63. $y=\frac{A x}{\sin (B x)}$

65. $\mathrm{y}=\frac{\mathrm{Ax}+\mathrm{B}}{\mathrm{Cx}+\mathrm{D}}$

In problems 67-71, (a) find $y$ ', (b) find the value(s) of $x$ so that $y^{\prime}=0$, and (c) find $y$ " Typically your answer in part (b) will contain As, Bs and (sometimes) Cs.

67. $\mathrm{y}=\mathrm{Ax}(\mathrm{B}-\mathrm{x})=\mathrm{ABx}-\mathrm{Ax}^{2}$

69. $\mathrm{y}=\mathrm{Ax}^{2}(\mathrm{~B}-\mathrm{x})=\mathrm{ABx}^{2}-\mathrm{Ax}^{3}$

71. $\mathrm{y}=\mathrm{Ax}^{3}+\mathrm{Bx}^{2}+\mathrm{C}$

Use the given differentiation patterns to differentiate the composite functions in problems 73-83. We have not derived the derivatives for these functions (yet), but if you are handed the derivative pattern for a function then you should be able to take derivatives of a composition involving that function.

Given:

$\begin{align*}\mathbf{D}(\arctan (x))=\frac{1}{1+x^{2}}, D(\arcsin (x))=\frac{1}{\sqrt{1-x^{2}}}, \quad \mathbf{D}(\ln (x))=\frac{1}{x}\end{align*}$

73. $\frac{\mathbf{d}}{\mathbf{d x}}\left(\arctan \left(x^{2}\right)\right)$

75. $\mathbf{D}\left(\arctan \left(\mathrm{e}^{\mathrm{x}}\right)\right)$

77. $\quad D\left(\arcsin \left(x^{3}\right)\right)$

79. $\frac{\mathbf{d}}{\mathbf{d t}}\left(\arcsin \left(\mathrm{e}^{\mathrm{t}}\right)\right)$

81. $\frac{\mathbf{d}}{\mathbf{d x}}(\ln (\sin (x)))$

83. $\frac{\mathbf{d}}{\mathrm{ds}}\left(\ln \left(\mathrm{e}^{\mathrm{s}}\right)\right)$

1. If $f(x)=x^{5}$ and $g(x)=x^{3}-7 x$, then $\operatorname{f ∘ g}(x)=\left(x^{3}-7 x\right)^{5}$

3. If $\mathrm{f}(\mathrm{x})=\mathrm{x}^{5 / 2}$ and $\mathrm{g}(\mathrm{x})=2+\sin (\mathrm{x})$, then $\operatorname{f ∘ g}(\mathrm{x})=$ $\sqrt{(2+\sin (x))^{5}}$ (The pair $f(x)=\sqrt{x}$ and $g(x)=(2+\sin (x))^{5}$ also work.)

5. If $f(x)=|x|$ and $g(x)=x^{2}-4$, then $\operatorname{f∘ g}(x)=\left|x^{2}-4\right|$

7. (1) $\mathrm{y}=\mathrm{u}^{5}, \mathrm{u}=\mathrm{x}^{3}-7 \mathrm{x}$

(2) $\mathrm{y}=\mathrm{u}^{4}, \mathrm{u}=\sin (3 \mathrm{x}-8)$

(3) $y=u^{5 / 2}, u=2+\sin (x)$

(4) $\mathrm{y}=1 / \sqrt{\mathrm{u}}, \mathrm{u}=\mathrm{x}^{2}+9$

(5) $\mathrm{y}=\mid \mathrm{u} \mathrm{I}, \mathrm{u}=\mathrm{x}^{2}-4$

(6) $\mathrm{y}=\tan (\mathrm{u}), \mathrm{u}=\sqrt{\mathrm{x}}$

9.

$\begin{array}{c|c|c|c|c|c|c}\mathrm{x} & \mathrm{f}(\mathrm{x}) & \mathrm{g}(\mathrm{x}) & \mathrm{f}^{\prime}(\mathrm{x}) & \mathrm{g}^{\prime}(\mathrm{x}) & (\mathrm{f} \circ \mathrm{g})(\mathrm{x}) & (\mathrm{f ∘ g})^{\prime}(\mathrm{x}) \\\hline-2 & 2 & -1 & 1 & 1 & \mathbf{1} & \mathbf{0} \\-1 & 1 & 2 & 0 & 2 & \mathbf{1} & \mathbf{2} \\0 & -2 & 1 & 2 & -1 & \mathbf{0} & \mathbf{1} \\1 & 0 & -2 & -1 & 2 & \mathbf{2} & \mathbf{2} \\2 & 1 & 0 & 1 & -1 & -\mathbf{2} & \mathbf{- 2}\end{array}$

11. $\mathrm{g}(2) \approx 2, \mathrm{~g}^{\prime}(2) \approx-1,(\operatorname{fog})(2)=\mathrm{f}(\mathrm{g}(2)) \approx \mathrm{f}(2) \approx 1$ $\mathrm{f}^{\prime}(\mathrm{g}(2)) \approx \mathrm{f}^{\prime}(2) \approx 0,(\text { fog })^{\prime}(2)=\mathrm{f}^{\prime}(\mathrm{g}(2)) \cdot \mathrm{g}^{\prime}(2) \approx 0$

13. $\mathrm{D}\left(\left(1-\frac{3}{\mathrm{x}}\right)^{4}\right)=4\left(1-\frac{3}{\mathrm{x}}\right)^{3}\left(\frac{3}{\mathrm{x}^{2}}\right)$

15. $\mathbf{D}\left(\frac{5}{\sqrt{2+\sin (\mathrm{x})}}\right)=5\left(\frac{-1}{2}\right)(2+\sin (\mathrm{x}))^{-3 / 2} \cos (\mathrm{x})=\frac{-5 \cos (\mathrm{x})}{2(2+\sin (\mathrm{x}))^{3 / 2}}$

17. $D\left(x^{2} \cdot \sin \left(x^{2}+3\right)\right)=x^{2}\left\{\cos \left(x^{2}+3\right)\right\}(2 x)+\left\{\sin \left(x^{2}+3\right)\right\}(2 x)=2 x\left\{x^{2} \cdot \cos \left(x^{2}+3\right)+\sin \left(x^{2}+3\right)\right\}$

19. $\operatorname{D}\left(\frac{7}{\cos \left(x^{3}-x\right)}\right)=\mathbf{D}\left(7 \sec \left(x^{3}-x\right)\right)=7\left(3 x^{2}-1\right) \cdot \sec \left(x^{3}-x\right) \cdot \tan \left(x^{3}-x\right)$

21. $D\left(e^{x}+e^{-x}\right)=e^{x}-e^{-x}$

23. $\mathrm{h}(\mathrm{t})=3-\cos (2 \mathrm{t})$ feet. $\quad$ (a) $\mathrm{h}(0)=2$ feet above the floor.

(b) $\mathrm{h}(\mathrm{t})=3-\cos (2 \mathrm{t})$ feet, $\mathrm{v}(\mathrm{t})=\mathrm{h}^{\prime}(\mathrm{t})=2 \sin (2 \mathrm{t}) \mathrm{ft} / \mathrm{sec}, \mathrm{a}(\mathrm{t})=\mathrm{v}^{\prime}(\mathrm{t})=4 \cos (2 \mathrm{t}) \mathrm{ft} / \mathrm{sec}^{2}$

(c) $\mathrm{K}=\frac{1}{2} \mathrm{mv}^{2}=\frac{1}{2} \mathrm{~m}(2 \sin (2 \mathrm{t}))^{2}=2 \mathrm{~m} \cdot \sin ^{2}(2 \mathrm{t}), \mathrm{dK} / \mathrm{dt}=8 \mathrm{~m} \cdot \sin (2 \mathrm{t}) \cdot \cos (2 \mathrm{t})$

25. $\mathrm{P}(\mathrm{h})=14.7 \mathrm{e}^{-0.0000385 \mathrm{~h}} .$ (a) $\mathrm{P}(0)=14.7$ psi (pounds per square inch), $\mathrm{P}(30,000) \approx 4.63 \mathrm{psi}$

(b) $10=14.7 \mathrm{e}^{-0.0000385 \mathrm{~h}}$ so $\frac{10}{14.7}=\mathrm{e}^{-0.0000385 \mathrm{~h}}$ and $\mathrm{h}=\frac{1}{-0.0000385} \ln \left(\frac{10}{14.7}\right) \approx 10,007 \mathrm{ft}$.

(c) $\mathrm{dP} / \mathrm{dh}=14.7(-0.0000385) \mathrm{e}^{-0.0000385 \mathrm{~h}} \mathrm{psi} / \mathrm{ft}$

At $\mathrm{h}=2,000$ feet, $\mathrm{dP} / \mathrm{dh}=14.7(-0.0000385) \mathrm{e}^{-0.0000385(2,000)} \mathrm{psi} / \mathrm{ft} \approx-0.000524 \mathrm{psi} / \mathrm{ft}$

Finally, $\mathrm{dP} / \mathrm{dt}=500(-0.000524) \approx-0.262$ psi/minute

(d) If the temperature is constant, then (pressure)(volume) is a constant (from physics!) so a decrease in pressure results in an increase in volume.

27. $\frac{\mathbf{d}}{\mathbf{d}} \sqrt{1+\cos ^{2}(\mathrm{z})}=\frac{2 \cos (\mathrm{z})\{-\sin (\mathrm{z})\}}{2 \sqrt{1+\cos ^{2}(\mathrm{z})}}=\frac{-\sin (2 \mathrm{z})}{2 \sqrt{1+\cos ^{2}(\mathrm{z})}}$

29. $\frac{d}{d x} \tan (3 x+5)=3 \cdot \sec ^{2}(3 x+5)$

31. $\mathbf{D}(\sin (\sqrt{x+1}))=\left\{\cos (\sqrt{x+1}) \frac{1}{2 \sqrt{x+1}}\right.$

33. $\frac{\mathbf{d}}{\mathrm{dx}}\left(\mathrm{e}^{\sin (\mathrm{x})}\right)=\mathrm{e}^{\sin (\mathrm{x})} \cdot \cos (\mathrm{x})$

35. $f(x)=\sqrt{x}$ so $\frac{\mathrm{d} \mathrm{f}(\mathrm{x})}{\mathrm{dx}}=\frac{1}{2 \sqrt{\mathrm{x}}} . \mathrm{x}(\mathrm{t})=2+\frac{21}{\mathrm{t}}$ so $\frac{\mathrm{d} \mathrm{x}(\mathrm{t})}{\mathrm{dt}}=-\frac{21}{\mathrm{t}^{2}}$.

At $\mathrm{t}=3, \mathrm{x}=9$ and $\frac{\mathrm{d} \mathrm{x}(\mathrm{t})}{\mathrm{dt}}=-\frac{21}{9}=-\frac{7}{3} \quad$ so $\frac{\mathrm{d}\left(\mathrm{f}_{\infty} \mathrm{x}\right)}{\mathrm{dt}}=\left(\frac{1}{2 \sqrt{9}}\right)\left(-\frac{7}{3}\right)=-\frac{7}{18}$

37. $f(x)=\tan ^{3}(x)$ so $\frac{\mathrm{d} \mathrm{f}(\mathrm{x})}{\mathrm{dx}}=3 \cdot \tan ^{2}(\mathrm{x}) \cdot \sec ^{2}(\mathrm{x}) . \mathrm{x}(\mathrm{t})=8$ so $\frac{\mathrm{d} \mathrm{x}(\mathrm{t})}{\mathrm{dt}}=0$

At $\mathrm{t}=3, \mathrm{x}=8$ and $\frac{\mathrm{d} \mathrm{x}(\mathrm{t})}{\mathrm{dt}}=0$ so $\frac{\mathrm{d}\left(\mathrm{f}_{\infty} \mathrm{x}\right)}{\mathrm{dt}}=0$

39. $f(x)=\frac{1}{77}(7 x-13)^{11}$

41. $f(x)=-\frac{1}{2} \cos (2 x-3)$

43. $f(x)=e^{\sin (x)}$

45. Then $-2 \sin (2 x)=2 \cos (x)\{-\sin (x)\}-2 \sin (x) \cdot \cos (x)$ or $\sin (2 x)=2 \sin (x) \cdot \cos (x)$.

47. $3 \cos (3 x)=3 \cos (x)-12 \sin ^{2}(x) \cdot \cos (x)$

so $\cos (3 x)=\cos (x)\left\{1-4 \sin ^{2}(x)\right\}=\cos (x)\left\{1-4+4 \cos ^{2}(x)\right\}=4 \cos ^{3}(x)-3 \cos (x)$

49. $y^{\prime}=3 \mathrm{Ax}^{2}+2 \mathrm{~B} \mathrm{x}$

51. $y^{\prime}=2 A x \cdot \cos \left(A x^{2}+B\right)$

53. $\mathrm{y}^{\prime}=\frac{\mathrm{Bx}}{\sqrt{\mathrm{A}+\mathrm{Bx}^{2}}}$

55. $y^{\prime}=B \cdot \sin (B x)$

57. $y^{\prime}=-2 A x \cdot \sin \left(A x^{2}+B\right)$

59. $y^{\prime}=x\left(B \cdot e^{B x}\right)+e^{B x}=(B x+1) \cdot e^{B x}$

61. $y^{\prime}=A \cdot e^{A x}+A \cdot e^{-A x}$

63. $y^{\prime}=\frac{A \cdot \sin (B x)-A x \cdot B \cdot \cos (B x)}{\sin ^{2}(B x)}$

65. $y^{\prime}=\frac{(C x+D) A-(A x+B) C}{(C x+D)^{2}}=\frac{A D-B C}{(C x+D)^{2}}$

67.

(a) $\mathrm{y}^{\prime}=\mathrm{AB}-2 \mathrm{Ax}$, (b) $\mathrm{x}=\frac{\mathrm{AB}}{2 \mathrm{~A}}=\frac{\mathrm{B}}{2}$, (c) $\mathrm{y}^{\prime \prime}=-2 \mathrm{~A}$

69.

(a) $\mathrm{y}^{\prime}=2 \mathrm{ABx}-3 \mathrm{Ax}^{2}=\mathrm{Ax} \cdot(2 \mathrm{~B}-3 \mathrm{x})$, (b) $\mathrm{x}=0,2 \mathrm{~B} / 3$

(c) $y^{\prime \prime}=2 \mathrm{AB}-6 \mathrm{Ax}$

(b) $\mathrm{x}=0, \frac{-2 \mathrm{~B}}{3 \mathrm{~A}}$

(c) $y^{\prime \prime}=6 A x+2 B$

71. (a) $\mathrm{y}^{\prime}=3 \mathrm{Ax}^{2}+2 \mathrm{~B} \mathrm{x}=\mathrm{x} \cdot(3 \mathrm{Ax}+2 \mathrm{~B})$,

(b) $\mathrm{x}=0, \frac{-2 \mathrm{~B}}{3 \mathrm{~A}}$,

(c) $\mathrm{y}^{\prime \prime}=6 \mathrm{Ax}+2 \mathrm{~B}$

73. $\frac{\mathbf{d}}{\mathrm{dx}}\left(\arctan \left(\mathrm{x}^{2}\right)\right)=\frac{2 \mathrm{x}}{1+\mathrm{x}^{4}}$

75. $\mathbf{D}\left(\arctan \left(\mathrm{e}^{\mathrm{x}}\right)\right)=\frac{1}{1+\left(\mathrm{e}^{\mathrm{x}}\right)^{2}} \cdot \mathrm{e}^{\mathrm{x}}=\frac{\mathrm{e}^{\mathrm{x}}}{1+\mathrm{e}^{2 \mathrm{x}}}$

77. $D\left(\arcsin \left(x^{3}\right)\right)=\frac{3 x^{2}}{\sqrt{1-x^{6}}}$

79. $\frac{\mathbf{d}}{\mathbf{d t}}\left(\arcsin \left(\mathrm{e}^{\mathrm{t}}\right)\right)=\frac{1}{\sqrt{1-\left(\mathrm{e}^{\mathrm{t}}\right)^{2}}} \cdot \mathrm{e}^{\mathrm{t}}=\frac{\mathrm{e}^{\mathrm{t}}}{\sqrt{1-\mathrm{e}^{2 \mathrm{t}}}}$

81. $\frac{\mathbf{d}}{\mathrm{dx}}(\ln (\sin (\mathrm{x})))=\frac{1}{\sin (\mathrm{x})} \cos (\mathrm{x})=\cot (\mathrm{x}) \quad$

83. $\frac{\mathbf{d}}{\mathrm{ds}}\left(\ln \left(\mathrm{e}^{\mathrm{s}}\right)\right)=\frac{1}{\mathrm{e}^{\mathrm{s}}} \cdot \mathrm{e}^{\mathrm{s}}=1$, or $\frac{\mathbf{d}}{\mathrm{ds}}\left(\ln \left(\mathrm{e}^{\mathrm{s}}\right)\right)=\frac{\mathbf{d}}{\mathrm{ds}}(\mathrm{s})=1$