\magnification1200
\def\newp#1.{\par\noindent{\bf #1.}}
\def\endp{\par\noindent\vrule width\hsize depth 0pt height .1pt}
\def\frac#1#2{{{\textstyle#1}\over{\textstyle#2}}}
\def\leaderfill{\leaders\hbox to .5em{\hss.\hss}\hfil}

\def\ans{\par\noindent\hbox to \hsize{\leaderfill}
\par\medskip}


\newp 13.
Find the area under the region
bounded by the curves $y=2^x$ and $y=5^x$, between $x=0$ and
$x=1$.
\ans
Since $5^x>2^x$ for $x>0$ the requested area is
$$\int_0^1\ 5^x\ -\ 2^x\ 
dx=\frac{5^x}{\ln 5}-\frac{2^x}{\ln 2}\Bigg\vert^1_0 =
\biggl(\frac{5}{\ln5}-\frac{2}{\ln2}\biggr)-
\biggl(\frac{1}{\ln5}-\frac{1}{\ln2}\biggr)=
\biggl(\frac{4}{\ln5}-\frac{1}{\ln2}\biggr)\ .
$$
\endp

\newp 14.
Evaluate the limit

$$\displaystyle\lim_{x \rightarrow 0^+}(\sin x )^{\tan x}$$
\ans
Let $\displaystyle y= \lim_{x\rightarrow 0^+}(\sin x)^{\tan x}$. Then 

$\displaystyle\ln y = \lim_{x\rightarrow 0^+}\tan x\ \ln(\sin x)=
\lim_{x\rightarrow 0^+}\frac{\ln(\sin x)}{\cot x}.$

When $x\rightarrow 0^+$, $\tan (x)\rightarrow 0$, 
$\cot (x)\rightarrow \infty$ and 
$\ln(\sin x))\rightarrow -\infty$.

This is an indeterminate form where we can apply L'Hospital's

rule, so

$\displaystyle\ln y=
 \lim_{x\rightarrow 0^+}\frac{\frac{\cos x}{\sin x}}{-\csc^2 x}=
\lim_{x\rightarrow 0^+}-\sin x \cos x=0$.

Since $\ln y=0$, $y=e^0=1$.

\endp


\newp 15.
Use logarithmic differentiation to
find the derivative of 
$$y= \frac{(x^2+1)^{2.2}(x+1)^{1.2}}{(x^3-1)^{0.3}}$$
\ans
This is a logarithmic differentiation problem, so start
by taking the log of both sides and simplifying the right
hand side.
$$\eqalign{%
\ln\> y=\ &
\ln\Biggl(
\frac{(x^2+1)^{2.2}(x+1)^{1.2}}{(x^3-1)^{0.3}}\Biggr)\cr
=\ &
\ln\bigl((x^2+1)^{2.2}\bigr)+
\ln\bigl((x+1)^{1.2}\bigr)-
\ln\bigl((x^3-1)^{0.3}\bigr)
\cr
=\ &
2.2\ln(x^2+1)+
1.2\ln(x+1)-
0.3 \ln(x^3-1)
\cr
}$$
Differentiating both sides with respect to $x$ we get
$$\eqalign{%
\frac{y^\prime}{y}\ =\ &
2.2\frac{2x}{x^2+1}+1.2\frac{1}{x+1}-0.3\frac{3x^2}{x^3-1}
\ =\ 
\frac{4\>x}{x^2+1}+\frac{1.2}{x+1}-\frac{0.9\>x^2}{x^3-1}\cr
}$$
Now solve for $y^\prime$
$$
\frac{dy}{dx}\ = \
\Biggl(
\frac{(x^2+1)^{2.2}(x+1)^{1.2}}{(x^3-1)^{0.3}}
\Biggr)\cdot\Biggl(
\frac{4\>x}{x^2+1}+\frac{1.2}{x+1}-\frac{0.9\>x^2}{x^3-1}
\Biggr)$$

\endp


\end