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Math 112 Solutions for 2.8, p. 186

1.
The graphs coincide if $ a$ is about 2.7.
2.
The graph looks a lot like that of $ \cos x$.
3.
(b) $ 2xe^{x}+x^{2}e^{x}$. (c) $ -e^{-x}$. (d) $ 2e^{2x}$. (f) $ e^{2x}(2\sin x+\cos x)$. (g) $ 2x\cos x-x^{2}\sin x$. (h) $ 2\cos^{2}x-2\sin^{2}x=2\cos 2x$. (j) $ 2\sin x\cos x$. (k) $ 2(\cos^{2}x-\sin^{2}x)=2\cos 2x$.
4.
(c) $ \dfrac{-3\sin x-4x^{2}\cos x-3x\cos x}{x^{2}\sin^{2}x}$. (f) $ 2^{x}(1+x\ln 2)$. (k) $ e^{2x}10^{x}(2+\ln 10)$.
5.
(a) $ e^{x}$. (c) $ 2^{x}(\ln 2)^{2}$. (d) $ 2e^{x}\cos x$. (e) $ \sec
x(\tan^{2}x+\sec^{2}x)$.
6.
The first six derivatives of $ \sin x$ are $ \cos x, -\sin x, -\cos
x, \sin x, \cos x, -\sin x$. Hence the fourth derivative of $ \cos x$ is the fifth derivative of $ \sin x$, which is $ \cos x$.
7.
$ y'=(e^{2x})'=(e^{x}\cdot e^{x})'=e^{x}\cdot e^{x}+e^{x}\cdot
e^{x}=2e^{2x}$; $ y''=(2e^{2x})'=2(e^{2x})'=2(2e^{2x})=2^{2}e^{2x}$; $ y'''=(2^{2}e^{2x})'=2^{2}(e^{2x})'=2^{2}(2e^{2x})=2^{3}e^{2x}$; ... $ y^{(n)}=2^{n}e^{2x}$.
8.

$\displaystyle \frac{d}{dx}(\cos x)$ $\displaystyle =\lim_{h\to 0}\frac{\cos(x+h)-\cos x}{h}=\lim_{h\to 0}\frac{\cos x\cos h-\sin x \sin h-\cos x}{h}$    
  $\displaystyle =\lim_{h\to 0}\left[ \frac{\cos h-1}{h}\cdot \cos x-\sin x\cdot\frac{\sin h}{h}\right] =0\cdot \cos x-\sin x\cdot 1=-\sin x.$    

9.

$\displaystyle \frac{d}{dx}(\tan x)$ $\displaystyle =\frac{d}{dx}\left(\frac{\sin x}{\cos x}\right)=\frac{\cos x\cdot\cos x-\sin x(-\sin x)}{\cos^{2}x}$    
  $\displaystyle =\frac{\cos^{2}x+\sin^{2}x}{\cos^{2}x}=\frac{1}{\cos^{2}x}=\sec^{2}x.$    

11.
$ \frac{d}{dx}(\sec x)=\frac{d}{dx}(1/\cos x)=\frac{-(-\sin
x)}{\cos^{2}x}=\frac{1}{\cos x}\cdot \frac{\sin x}{\cos x}=\sec x\tan x$.

13.
(f) $ v=e^{-t}(-\sin t+\cos t), a=-2e^{-t}\cos t, v(0)=1, a(0)=-2$.

14.
(a) $ v(t)=1-\cos t, a(t)=\sin t$. (b) $ v(t)=0, t>0\Rightarrow
\cos t=1\Rightarrow t=2\pi$. (c) $ a(t)=0, t>0\Rightarrow \sin t=0\Rightarrow
t=\pi; v(\pi)=2$.

15.
(c) $ y=\pi-x$. (f) $ y=\frac{2}{e}-\frac{1}{e}x$.

18.
$ \frac{d}{dx}(\sinh
x)=\frac{d}{dx}[\frac{1}{2}(e^{x}-\frac{1}{e^{x}})]
=\frac{1}{2}(e^{x}-\frac{-e^{x}}{e^{2x}})=\frac{1}{2}(e^{x}+e^{-x})=\cosh x$, and similarly for $ \cosh x$.




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Jason Grout 2003-02-10