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\SN{6.2}

\pn1. \hskip 6pt a) $\ln 0.75=\ln\frac 34=\ln3-\ln4=\ln3-\ln2^2=\ln3-2\ln2$.
\itemitem{b)} $\ln(\frac 49)=\ln\bigl((\frac 23)^2\bigr)=
2(\ln\frac 23)=2(\ln2-\ln3)$.
\itemitem{c)}$\ln\frac 12=-\ln 2$.
\itemitem{d)} $\ln\root3\of9=\ln 9^{1/3}=\frac 13\cdot\ln 9=
\frac 13\ln(3^2)=\frac 23\ln3$.
\itemitem{e)} $\ln 3\sqrt2=\ln(3^{1/2})=\frac 12\cdot\ln 3$.
\itemitem{f)} $\ln \sqrt{13.5}=\ln\sqrt{\frac{27}2}=
\ln\bigl((\frac{27}2)^{1/2}\bigr)=\frac 12\cdot\ln(\frac{27}2)=
\frac 12\bigl(\ln 27 - \ln 2\bigr)=
\frac 12\bigl(\ln 3^3-\ln2\bigr)=
\frac 12\bigl(3\ln3-\ln2\bigr)$.

\def\dd #1#2{\frac{d#1}{d#2}}

\pn5. $\dd yx=\frac{\frac{d(3x)}{dx}}{3x}=\frac 3{3x}=\frac 1x$.

\pn11.$\dd y\theta=\frac{\frac{d(\theta+1)}{dx}}{\theta+1}
=\frac 1{\theta+1}$.

\pn15.$\dd yt= \dd tt \cdot(\ln t)^2+ t\cdot\frac {d(\ln t)^2}{dt}=
(\ln t)^2+t\cdot\bigl(2(\ln t)^{2-1}\cdot \frac{d(\ln t)}{dt}\bigr)=
(\ln t)^2+2t(\ln t)\frac 1t=(\ln t)^2+2\ln t$.
{\sl Product Rule, then Power Rule.}
\pn19. $\dd yt=\frac{ (\frac{d(\ln t}{dt}\cdot t)-(\ln t\cdot\frac{d(t)}{dt})}
{t^2}=
\frac{ (\frac 1t )t-(\ln t)}{t^2}=\frac{1-\ln t}{t^2}$.
{\sl Quotient Rule.}

\pn21. $\dd yx=\frac{\bigl(\frac{d(\ln x)}{dx}(1+\ln x)\bigr)-
\bigl((\ln x)(\frac{d(1+\ln x)}{dx}}{(1+\ln x)^2}=
\ifdistill\hfill\penalty-10000 \fi%
\frac{ \bigl((\frac 1x)(1+\ln x)\bigr)-\bigl((\ln x)(\frac 1x)\bigr)}
{(1+\ln x)^2}=
\frac{\frac 1x}{(1+\ln x)^2}=\frac 1{x(1+\ln x)^2}$.
\ifdistill\hfill\penalty-10000 \fi%
{\sl Quotient Rule again.}

\pn25. $\dd y\theta =
\frac{d(\theta)}{d\theta}\bigl(\sin(\ln\theta)+\cos(\ln\theta)\bigr)+
\ifdistill\hfill\penalty-10000 \fi%
\theta\Bigl(\frac{d\bigl(\sin(\ln\theta)+\cos(\ln\theta)\bigr)}{d\theta}\Bigr)=
\bigl(\sin(\ln\theta)+\cos(\ln\theta)\bigr)+
\theta\bigl(\frac{d(\sin(\ln\theta))}{d\theta}+
\frac{d(\cos(\ln\theta))}{d\theta}\bigr)=
\bigl(\sin(\ln\theta)+\cos(\ln\theta)\bigr)+
\theta\bigl(\frac{d(\sin(\ln\theta))}{d\theta}+
\frac{d(\cos(\ln\theta))}{d\theta}\bigr)=
\bigl(\sin(\ln\theta)+\cos(\ln\theta)\bigr)+
\theta\bigl(\cos(\ln\theta)\frac{d(\ln\theta)}{d\theta}
-\sin(\ln\theta)\frac{d(\ln\theta)}{d\theta}\bigr)=
\bigl(\sin(\ln\theta)+\cos(\ln\theta)\bigr)+
\theta\bigl(\cos(\ln\theta)\frac{1}{\theta}+
-\sin(\ln\theta)\frac{1}{\theta}\bigr)=
2\cos(\ln\theta)$.
{\sl Product Rule, Sum Rule, Chain Rule (twice).}

\pn37. Find $\dd yx$: $\ln y=\ln\sqrt{x(x+1)}=
\frac 12\ln\bigl(x(x+1)\bigr)=
\frac 12\bigl(\ln x+\ln(x+1)\bigr)$, so
$\frac{\frac{dy}{dx}}{y}=\frac 12\bigl(\frac 1x+\frac 1{1+x}\bigr)$,
so
$\dd yx=\bigl(\sqrt{x(x+1)}\>\bigr)\bigl(
\frac 12\bigl(\frac 1x+\frac 1{1+x}\bigr)\bigr)$.

\pn49. $\ln y=\ln\root3\of{\frac{x(x-2)}{x^2+1}}=
\frac 13\bigl(\ln x+\ln (x-2)-\ln(x^2+1)\bigr)$, so
$\frac{\frac{dy}{dx}}{y}=\frac 13\bigl(\frac 1x+\frac 1{x-2}-
\frac{\frac{d(x^2+1)}{dx}}{x^2+1}\bigr)=
\frac 13\bigl(\frac 1x+\frac 1{x-2}-
\frac{2x}{x^2+1}\bigr)$, so
$\dd yx=\frac 13\root3\of{\frac{x(x-2)}{x^2+1}}
\Bigl(\frac 1x+\frac 1{x-2}-
\frac{2x}{x^2+1}\Bigr)$.

\pn53. $\int \frac {2y\ dy}{y^2-25}$.
Substitute $u=y^2-25$: $du=2y\ dy$, so our integral becomes
$\int \frac {du}{u}=\ln\vert u\vert + C=
\ln\vert y^2-25\vert +C$.

\pn57. $\int_1^2 \frac{2\ln x}{x}\> dx$. Substitute $u=\ln x$: 
$du=\frac 1x\>dx$.
Hence our integral becomes
$\int_{\ln 1}^{\ln 2} 2u\ du=u^2\Big\vert_{0}^{\ln2}=(\ln 2)^2-(0)^2=
(\ln2)^2$.
Note $(\ln 2)^2\neq\ln 4$! It is true that $(\ln 2)^2=\ln2\cdot \ln2=
\ln(2^{\ln2})$.

\pn59. $\int_2^4 \frac{dx}{x(\ln x)^2}$. Substitute $u=\ln x$:
$du=\frac 1x\ dx$ so our integral becomes
$\int_{\ln 2}^{\ln 4} \frac {du}{u^2}=
-u^{-1}\Big\vert_{\ln2}^{2\ln2}=
-\frac {1}{2\ln2}+\frac 1{\ln2}=\frac {1}{2\ln2}$.
Note $2\ln2=\ln4$.

\pn61. $\int \frac{3\sec^2 t}{6+3\tan t}\> dt$.
Substitute $u=6+3\tan t$: $du=3\sec^2 t\ dt$ so our
integral becomes
$\int \frac{du}{u}=\ln\vert u\vert+C=
\ln\vert 6+3\tan t\vert + C$.



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