\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}