\begin{frame} \frametitle{2nd Midterm Exam - Review} \begin{exampleblock}{} Find the derivative of $f(x)$ when $0 < x < 5$: \begin{align*} f(x) = \frac{|x|}{\sqrt{5^2 - x^2}} \end{align*} \pause If $0 < x$ then $|x| = x$. Then \begin{align*} f(x) &= \frac{x}{\sqrt{5^2 - x^2}}\\ \mpause[1]{f'(x)} &\mpause[1]{=} \mpause[2]{\frac{1\cdot \sqrt{5^2 - x^2} - x\cdot \frac{1}{2}(5^2 - x^2)^{-\frac{1}{2}}\cdot (-2x)}{(\sqrt{5^2 - x^2})^2}} \\ &\mpause[3]{= \frac{(5^2 - x^2) + x^2}{(\sqrt{5^2 - x^2})^3}} \mpause[4]{= \frac{25}{(5^2 - x^2)^{\frac{3}{2}}}} \end{align*} \pause\pause\pause\pause\pause What is the left-hand derivative at $0$? \pause Then $x < 0$, thus $|x| = -x$. \pause \begin{align*} \text{left-hand derivative at $0$} = -1/5 \end{align*} \end{exampleblock} \end{frame}