\begin{frame} \frametitle{Derivatives of Trigonometric Functions} We will prove that \begin{block}{} \begin{malign} \frac{d}{dx} \sin(x) \;=\; \cos(x) \end{malign} \end{block} We have \begin{talign} \hspace{.5cm}\frac{d}{dx} \sin(x) &= \sin x \cdot \lim_{h\to 0}\frac{\cos h - 1}{h} + \cos x \hspace{5cm} \\ &\mpause[1]{= \sin x \cdot \lim_{h\to 0}\left( \frac{\cos h - 1}{h} \cdot \frac{\cos h +1}{\cos h + 1} \right) + \cos x } \\ &\mpause[2]{= \sin x \cdot \lim_{h\to 0}\left( \frac{(\cos h)^2 - 1}{h(\cos h + 1)} \right) + \cos x } \\ &\mpause[3]{= \sin x \cdot \lim_{h\to 0}\left( \frac{-(\sin h)^2}{h(\cos h + 1)} \right) + \cos x } \\ &\mpause[4]{= \sin x \cdot \left( \lim_{h\to 0} \frac{\sin h}{h} \cdot \lim_{h\to 0} \frac{-\sin h}{\cos h + 1} \right) + \cos x } \\ &\mpause[5]{= \sin x \cdot 1 \cdot 0 + \cos x } \mpause[6]{= \cos x } \end{talign} \vspace{10cm} \end{frame}