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Section 5.4 #43 : Compute the derivative of $g(t)=(e^{-t}+e^t)^3$.
Solution: Calculate using the power rule and chain rule (twice!) to get
$$\begin{array}{ll}
\dfrac{\mathrm{d}}{\mathrm{d}t} (e^{-t}+e^t)^3 &= 3(e^{-t}+e^t)^2 \dfrac{\mathrm{d}}{\mathrm{d}t} [e^{-t}+e^t] \\
&= 3(e^{-t}+e^t)^2 \left( e^{-t} \dfrac{\mathrm{d}}{\mathrm{d}t} [-t] + e^t \right) \\
&= 3(e^{-t}+e^t)^2(e^t-e^{-t}).
\end{array}$$
Section 5.4 #46: Compute the derivative of $y=\ln \left( \dfrac{1+e^x}{1-e^x} \right)$.
Solution: Using the chain rule and the quotient rule, compute
$$\begin{array}{ll}
\dfrac{\mathrm{d}}{\mathrm{d}x} \ln \left( \dfrac{1+e^x}{1-e^x} \right) &= \dfrac{1}{\frac{1+e^x}{1-e^x}} \dfrac{\mathrm{d}}{\mathrm{d}x} \left[ \dfrac{1+e^x}{1-e^x} \right] \\
&= \dfrac{1-e^x}{1+e^x} \dfrac{(1-e^x)e^x - (1+e^x)(-e^x)}{(1-e^x)^2} \\
&= \dfrac{e^x-e^{2x}+e^x + e^{2x}}{(1+e^x)(1-e^x)} \\
&= \dfrac{2e^x}{(1+e^x)(1-e^x)} \\
&= \dfrac{2e^x}{1-e^{2x}}.
\end{array}$$
Section 5.4 #99: Compute $\displaystyle\int \dfrac{e^{-x}}{1+e^{-x}} \mathrm{d}x$.
Solution: Let $u=1+e^{-x}$ so that $\mathrm{d}u=-e^{-x} \mathrm{d}x$, or equivalently, $-\mathrm{d}u=e^{-x} \mathrm{d}x$. Now calculate
$$\begin{array}{ll}
\displaystyle\int \dfrac{e^{-x}}{1+e^{-x}} \mathrm{d}x &= -\displaystyle\int \dfrac{1}{u} \mathrm{d}u \\
&=-\ln(|u|) + C \\
&= - \ln (|1+e^{-x}|) + C \\
&= -\ln(1+e^{-x})+C.
\end{array}$$
Section 5.4 #103: Compute $\displaystyle\int \dfrac{e^x + e^{-x}}{e^x-e^{-x}} \mathrm{d}x$.
Solution: Let $u=e^x-e^{-x}$ so that $\mathrm{d}u=e^x+e^{-x} \mathrm{d}x$. Now compute
$$\begin{array}{ll}
\displaystyle\int \dfrac{e^x+e^{-x}}{e^x-e^{-x}} \mathrm{d}x &= \displaystyle\int \dfrac{1}{u} \mathrm{d}u \\
&= \ln (|u|) + C \\
&= \ln (e^x - e^{-x}) + C \\
&= \ln \left( \dfrac{e^{2x}-1}{e^x} \right) + C \\
&= \ln (e^{2x}-1) - \ln(e^x) + C \\
&= \ln(e^{2x}-1)-x+C.
\end{array}$$
(note: full credit for making it to any lines after the $\ln(|u|)+C$ line)
Section 5.5 #40: Calculate the derivative of $y=6^{3x-4}$.
Solution: Note that
$$6^{3x-4} = \exp \left( \ln \left( 6^{3x-4} \right) \right) = e^{(3x-4)\ln(6)}.$$
Therefore compute using the chain rule,
$$\begin{array}{ll}
\dfrac{\mathrm{d}}{\mathrm{d}x} 6^{3x-4} &= \dfrac{\mathrm{d}}{\mathrm{d}x} e^{(3x-4)\ln(6)} \\
&= e^{(3x-4)\ln(6)} \dfrac{\mathrm{d}}{\mathrm{d}x} \left[ (3x-4)\ln(6) \right] \\
&= 3\ln(6) 6^{3x-4}.
\end{array}$$
Section 5.5 #78: Calculate $\displaystyle\int 2^{\sin(x)} \cos(x) \mathrm{d}x$.
Solution: Let $u=\sin(x)$ so that $\mathrm{d}u=\cos(x) \mathrm{d}x$. Also note that $2^u = e^{u\ln(2)}$. Now compute
$$\begin{array}{ll}
\displaystyle\int 2^{\sin(x)} \cos(x) \mathrm{d}x &= \displaystyle\int 2^u \mathrm{d}u \\
&= \displaystyle\int e^{u\ln(2)} \mathrm{d}u.
\end{array}$$
At this point it is advantageous to do another substitution. Let $w=u\ln(2)$ so that $\mathrm{d}w=\ln(2) \mathrm{d}u$, or $\dfrac{1}{\ln(2)} \mathrm{d}w = \mathrm{d}u$. This yields
$$\begin{array}{ll}
\displaystyle\int e^{u\ln(2)} \mathrm{d}u &= \dfrac{1}{\ln(2)} \displaystyle\int e^w \mathrm{d}w \\
&=\dfrac{1}{\ln(2)} e^w + C \\
&\stackrel{w=u\ln(2)}{=} \dfrac{1}{\ln(2)} e^{u\ln(2)} + C \\
&= \dfrac{1}{\ln(2)} 2^u + C \\
&\stackrel{u=\sin(x)}{=} \dfrac{1}{\ln(2)} 2^{\sin(x)} + C.
\end{array}$$
