\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} \pause We have \begin{talign} \hspace{.5cm}\frac{d}{dx} \sin(x) &= \mpause[1]{ \lim_{h\to 0} \frac{\sin(x+h)- \sin(x)}{h} } \hspace{5cm} \\ &\mpause[2]{= \lim_{h\to 0} \frac{\sin x \cos h + \cos x \sin h - \sin(x)}{h} } \\ &\mpause[3]{= \lim_{h\to 0} \left[ \frac{\sin x \cos h - \sin(x)}{h} + \frac{\cos x \sin h}{h} \right] } \\ &\mpause[4]{= \lim_{h\to 0} \left[ \sin x\frac{\cos h - 1}{h} + \cos x\frac{\sin h}{h} \right] } \\ &\mpause[5]{= \sin x \cdot \lim_{h\to 0}\frac{\cos h - 1}{h} + \cos x \cdot \alert<8->{\underbrace{\lim_{h\to 0}\frac{\sin h}{h}}_{1}} } \end{talign} \vspace{10cm} \end{frame}