**Exercise:** Find the derivative. Simplify where possible.

- h(x)=\ln(\cosh x)
- y=\sinh^{-1}(\tan x)

- h'(x)=\sinh x / \cosh x=\tanh x
- y'=\sec^2 x/\sqrt{1+\tan^2x}=\sec^2 x/\sqrt{\sec^2 x}=|\sec x|

**Exercise:** Evaluate \lim_{x\to\infty}\frac{\sinh x}{e^x}

Since \frac{\sinh x}{e^x}=\frac{e^x-e^{-x}}{2e^x}=\frac{1-e^{-2x}}{2}, \lim_{x\to\infty}\frac{\sinh x}{e^x}=\frac{1}{2}.

**Exercise:** Evaluate the integral.

- \int \ln(2x+1)dx
- \int_1^3 r^3\ln rdr

- Let u=2x+1. Then \begin{aligned}\int \ln(2x+1)dx & = \frac{1}{2}\int \ln(u)du\\& = \frac{1}{2}(u\ln u-\int u/u du)\\& = \frac{1}{2}(u\ln u-u)+C\\ & = \frac{1}{2}((2x+1)\ln(2x+1)-(2x+1))+C\end{aligned}.
- \begin{aligned}\int_1^3 r^3\ln rdr & = \int_1^3 \ln rd(\frac{r^4}{4})\\& = \ln r(\frac{r^4}{4})|_1^3-\int_1^3(\frac{r^4}{4})/rdr \\ & = \frac{81}{4}\ln 3-\frac{r^4}{16}|_1^3 \\ & = \frac{81}{4}\ln 3-5\end{aligned}.