\begin{frame} \frametitle{Differentiation Rules: Product Rule} \begin{exampleblock}{} Let $f(x) = xe^x$. Find $f'(x)$. \pause \begin{talign} f'(x) &\mpause[1]{= \frac{d}{dx} (x \cdot e^x)} \\ &\mpause[2]{= x \frac{d}{dx} (e^x) + e^x \frac{d}{dx} (x)} \\ &\mpause[3]{= x e^x + e^x} \\ &\mpause[4]{= (x+1) e^x} \end{talign} \end{exampleblock} \pause\pause\pause\pause\pause \begin{exampleblock}{} Let $f(x) = xe^x$. Find the $n$-th derivative $f^{(n)}(x)$. \pause \begin{talign} f''(x) &\mpause[1]{= \frac{d}{dx} (x e^x + e^x)} \mpause[2]{= (x+1) e^x + e^x} \mpause[3]{= (x+2) e^x} \\ \mpause[4]{f'''(x)} &\mpause[5]{= \frac{d}{dx} (x e^x + 2e^x)} \mpause[6]{= (x+1) e^x + 2e^x} \mpause[7]{= (x+3) e^x} \end{talign} \pause\pause\pause\pause\pause\pause\pause\pause Thus obviously we have \begin{talign} f^{(n)}(x) &= (x+n) e^x \end{talign} \end{exampleblock} \end{frame}