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\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}