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\begin{frame}
\frametitle{Differentiation Rules: Quotient Rule}

Assume that $f$ and $g$ are differentiable at $x$, and define
\begin{talign}
h(x) &= \frac{f(x)}{g(x)}
&
\mpause[1]{
\begin{aligned}
\Delta h &= h(x+\Delta x) - h(x) \\
\Delta f &= f(x+\Delta x) - f(x) \\
\Delta g &= g(x+\Delta x) - g(x)
\end{aligned}
}
\end{talign}
\pause\pause
We try to find the derivative of $h$ at $x$:
\begin{talign}
\Delta h &= h(x+\Delta x) - h(x)
\mpause[1]{= \frac{f(x + \Delta x)}{g(x + \Delta x)} - \frac{f(x)}{g(x)}}
\mpause[2]{= \frac{f(x) + \Delta f}{g(x) + \Delta g} - \frac{f(x)}{g(x)}} \\
&\mpause[3]{= \frac{(f(x) + \Delta f)\cdot g(x) - (g(x) + \Delta g)\cdot f(x)}{(g(x) + \Delta g)\cdot g(x)}}
\mpause[4]{= \frac{g(x) \Delta f - f(x) \Delta g}{(g(x) + \Delta g)\cdot g(x)}} \\[1ex]
%
\mpause[5]{\alert<13->{h'(x)} }&\mpause[6]{= \lim_{\Delta x \to 0} \frac{\Delta h}{\Delta x}}
\mpause[7]{= \lim_{\Delta x \to 0} \frac{\frac{g(x) \Delta f - f(x) \Delta g}{(g(x) + \Delta g)\cdot g(x)}}{\Delta x}}
\mpause[8]{= \lim_{\Delta x \to 0} \frac{g(x) \frac{\Delta f}{\Delta x} - f(x) \frac{\Delta g}{\Delta x}}{(g(x) + \Delta g)\cdot g(x)}} \\
&\mpause[9]{= \frac{g(x) \lim_{\Delta x \to 0}\frac{\Delta f}{\Delta x} - f(x) \lim_{\Delta x \to 0}\frac{\Delta g}{\Delta x}}{\lim_{\Delta x \to 0} (g(x) + \Delta g)\cdot g(x)}}
\mpause[10]{= \alert<13->{\frac{g(x) f'(x) - f(x) g'(x)}{g(x)^2}} }
\end{talign}
\end{frame}