\begin{frame} \frametitle{Implicit Differentiation} \begin{block}{} We can use \emph{implicit differentiation}: \begin{itemize} \pause \item differentiate both sides of the equation w.r.t. $x$, and \pause \item then solve for $y'$, that is, for $\frac{dy}{dx}$ \end{itemize} \end{block} \pause \begin{exampleblock}{} We differentiate $x^2 + y^2 = 25$ implicitly. \pause We have \begin{talign} &\frac{d}{dx}(x^2 + y^2) = \frac{d}{dx} 25 \\[.5ex] &\mpause[1]{ \frac{d}{dx} x^2 + \frac{d}{dx} y^2 = 0 } \\[.5ex] &\mpause[2]{ 2x + \frac{d}{dx} y^2 = 0 } \hspace{1cm} \mpause[3]{ \text{\textcolor{gray}{$y$ is a function of $x$ $\implies$ chain rule}} }\\[.5ex] &\mpause[4]{ 2x + \frac{d}{dy}(y^2) \frac{d}{dx}y = 0 } \\[.5ex] &\mpause[5]{ 2x + 2y \frac{d}{dx}y = 0 } \mpause[6]{\;\;\implies\;\; \frac{d}{dx}y = -\frac{x}{y} } \mpause[7]{\;\;\implies\;\; \frac{dy}{dx} = -\frac{x}{y} } \end{talign} \end{exampleblock} \vspace{10cm} \end{frame}