Practice Problems

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 $\frac{\mathbf{d}}{\mathrm{ds}}\left(\ln \left(\mathrm{e}^{\mathrm{s}}\right)\right)=\frac{\mathbf{d}}{\mathrm{ds}}(\mathrm{s})=1$