Practice Problems

1.

$\begin{aligned}&\mathbf{D}\left(\mathrm{f}^{2}(\mathrm{x})\right)=2 \cdot \mathrm{f}^{1}(\mathrm{x}) \cdot \mathrm{f}^{\prime}(\mathrm{x}) . \text { At } \mathrm{x}=1, \mathbf{D}\left(\mathrm{f}^{2}(\mathrm{x})\right)=2(2)(3)=12 \\&\mathbf{D}\left(\mathrm{f}^{5}(\mathrm{x})\right)=5 \cdot \mathrm{f}^{4}(\mathrm{x}) \cdot \mathrm{f}^{\prime}(\mathrm{x}) . \text { At } \mathrm{x}=1, \mathrm{D}\left(\mathrm{f}^{5}(\mathrm{x})\right)=5(2)^{4}(3)=240 \\&\mathbf{D}\left(\mathrm{f}^{1 / 2}(\mathrm{x})\right)=(1 / 2) \cdot \mathrm{f}^{-1 / 2}(\mathrm{x}) \cdot \mathrm{f}^{\prime}(\mathrm{x}) . \text { At } \mathrm{x}=1, \mathbf{D}\left(\mathrm{f}^{1 / 2}(\mathrm{x})\right)=(1 / 2)(2)^{-1 / 2}(3)=\frac{3}{2 \sqrt{2}}=\frac{3 \sqrt{2}}{4}\end{aligned}$

3. 

$\begin{array}{llllll}x & f(x) & f^{\prime}(x) & D\left(f^{2}(x)\right) & D\left(f^{3}(x)\right) & D\left(f^{5}(x)\right) \\1 & 1 & -1 & -2 & -3 & -5 \\3 & 2 & -3 & -12 & -36 & -240\end{array}$

5. $\quad \mathrm{f}^{\prime}(\mathrm{x})=5 \cdot(2 \mathrm{x}-8)^{4} \cdot(2)$

7. $\quad \mathrm{f}^{\prime}(\mathrm{x})=\mathrm{x} \cdot 5 \cdot(3 \mathrm{x}+7)^{4} \cdot 3+1 \cdot(3 \mathrm{x}+7)^{5}=(3 \mathrm{x}+7)^{4}\{15 \mathrm{x}+(3 \mathrm{x}+7)\}=(3 \mathrm{x}+7)^{4} \cdot(18 \mathrm{x}+7)$

9. $\quad f^{\prime}(x)=\frac{1}{2}\left(x^{2}+6 x-1\right)^{-1 / 2} \cdot(2 x+6)=\frac{x+3}{\sqrt{x^{2}+6 x-1}}$

11. (a) graph $h(t) = 3-2 \sin (\mathrm{t})$

(b) When $t = 0, h(0) = 3$ feet.

(c) Highest $=5$ feet above the floor. Lowest $=1$ foot above the floor.

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

(e) This spring oscillates forever. The motion of a real spring would "damp out" due to friction.

13. $\quad \mathrm{K}=\frac{1}{2} \mathrm{mv}^{2} \quad$ (a) If $\mathrm{h}(\mathrm{t})=5 \mathrm{t}$, then $\mathrm{v}(\mathrm{t})=\mathrm{h}^{\prime}(\mathrm{t})=5 .$ Then $\mathrm{K}(1)=\mathrm{K}(2)=\frac{1}{2} \mathrm{~m}(5)^{2}=12.5 \mathrm{~m}$.

(b) If $h(t)=t^{2}$, then $v(t)=h^{\prime}(t)=2 t$ so $v(1)=2$ and $v(2)=4$. Then $K(1)=\frac{1}{2} m(2)^{2}=2 m$ and $\mathrm{K}(2)=\frac{1}{2} \mathrm{~m}(4)^{2}=8 \mathrm{~m}$

15. $\frac{\mathrm{df}}{\mathrm{dx}}=\mathrm{x} \cdot \mathbf{D}(\sin (\mathrm{x}))+\sin (\mathrm{x}) \cdot \mathbf{D}(\mathrm{x})=\mathrm{x} \cdot \cos (\mathrm{x})+\sin (\mathrm{x})$

17. $\mathrm{f}^{\prime}(\mathrm{x})=\mathrm{e}^{\mathrm{x}}-\sec (\mathrm{x}) \cdot \tan (\mathrm{x})$

19. $\mathrm{f}^{\prime}(\mathrm{x})=-\mathrm{e}^{-\mathrm{x}}+\cos (\mathrm{x})$

21. $\mathrm{f}^{\prime}(\mathrm{x})=7(\mathrm{x}-5)^{6}(1)$ so $\mathrm{f}^{\prime}(4)=7(-1)^{6}(1)=7$. Then $\mathrm{y}-(-1)=7(\mathrm{x}-4)$ and $\mathrm{y}=7 \mathrm{x}-29$.

23. $\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{2}\left(25-\mathrm{x}^{2}\right)^{-1 / 2}(-2 \mathrm{x})=\frac{-\mathrm{x}}{\sqrt{25-\mathrm{x}^{2}}}$ so $\mathrm{f}^{\prime}(3)=\frac{-3}{4}$. Then $\mathrm{y}-4=-\frac{3}{4}(\mathrm{x}-3)$ or $3 \mathrm{x}+4 \mathrm{y}=25$

25. $\mathrm{f}^{\prime}(\mathrm{x})=5(\mathrm{x}-\mathrm{a})^{4}(1)$ so $\mathrm{f}^{\prime}(\mathrm{a})=5(\mathrm{a}-\mathrm{a})^{4}(1)=0 .$ Then $\mathrm{y}-0=0(\mathrm{x}-\mathrm{a})$ or $\mathrm{y}=0$

27. $f^{\prime}(x)=e^{x}$ so $f^{\prime}(3)=e^{3}$. Then $y-e^{3}=e^{3}(x-3)$.

$x$-intercept $(y=0): 0-e^{3}=e^{3}(x-3)$ so $-1=x-3$ and $x=2$

$\left(y-\right.$ intercept $(x=0): y-e^{3}=e^{3}(0-3)$ so $y=-3 e^{3}+e^{3}=-2 e^{3}$

At $\left(p, e^{p}\right), f^{\prime}(p)=e^{p}$ so $y-e^{p}=e^{p}(x-p)$

$x$-intercept $(y=0): 0-e^{p}=e^{p}(x-p)$ so $-1=x-p$ and $x=p-1$

29. $\mathrm{f}^{\prime}(\mathrm{x})=-\sin (\mathrm{x}), \mathrm{f} "(\mathrm{x})=-\cos (\mathrm{x})$

31. $\quad \mathrm{f}^{\prime}(\mathrm{x})=\mathrm{x}^{2} \cos (\mathrm{x})+2 \mathrm{x} \sin (\mathrm{x})$,

$\begin{align*}\mathrm{f}^{\prime \prime}(\mathrm{x})=-\mathrm{x}^{2} \sin (\mathrm{x})+2 \mathrm{x} \cos (\mathrm{x})+2 \mathrm{x} \cos (\mathrm{x})+2 \sin (\mathrm{x})=-\mathrm{x}^{2} \sin (\mathrm{x})+4 \mathrm{x} \cos (\mathrm{x})+2 \sin (\mathrm{x})\end{align*}$

33. $f^{\prime}(x)=e^{x} \cdot \cos (x)-e^{x} \cdot \sin (x), f^{\prime \prime}(x)=-2 e^{x} \cdot \sin (x)$

35. $q^{\prime}=$ linear, $q^{\prime \prime}=$ constant, $q^{\prime \prime \prime}=q^{(4)}=q^{(5)}=\ldots=0$

37. $\mathrm{P}^{(\mathrm{n})}=$ constant, $\mathrm{p}^{(\mathrm{n}+1)}=0$

39. $f(x)=5 e^{x}$

41. $f(x)=\left(1+e^{x}\right)^{5}$

43. No.

$\lim\limits_{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}=\lim\limits_{h \rightarrow 0} \frac{(0+h) \cdot \sin \left(\frac{1}{0+h}\right)-0}{h}=\lim\limits_{h \rightarrow 0} \sin \left(\frac{1}{h}\right)$ which does not exist.

(To see that this last limit does not exist, graph $\sin (1 / \mathrm{h})$ for $-1 \leq \mathrm{h} \leq 1$, or evalaute $\sin (1 / \mathrm{h})$ for lots of small values of $h, e . g .$, h $0.1,0.01,0.001, \ldots$ )

45. $\left(1+\frac{1}{\mathrm{X}}\right)^{\mathrm{x}} \approx 2.718 \ldots=\mathrm{e}$ when $\mathrm{x}$ is large.

47. (a) $s_{2}=2.5, s_{3} \approx 2.67, s_{4} \approx 2.708, \mathrm{~s}_{5} \approx 2.716, \mathrm{~s}_{6} \approx 2.718, \mathrm{~s}_{7} \approx 2.71825, \mathrm{~s}_{8} \approx 2.718178$

(b) They are approaching $e$.