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If $f$ and $g$ are functions that are differentiable at $x$, then the derivative of their product exists and is given by
$$\frac{d}{dx} \left[ \frac{f(x)}{g(x)} \right]= \frac{g(x) f'(x) + f(x) g'(x)}{[g(x)]^2}$$Proof:
After appealing to the limit definition of the derivative and collapsing the resulting complex fraction, we add a well-chosen value of zero, $-g(x)f(x)+g(x)f(x)$, to the numerator:
$$\begin{array}{rcl} \dfrac{d}{dx} \left[ \dfrac{f\,(x)}{ g(x)} \right] & = & \displaystyle{\lim_{h \rightarrow 0} \dfrac{ \dfrac{f(x+h)}{g(x+h)} - \dfrac{f(x)}{g(x)}}{h}}\\\\ & = & \displaystyle{\lim_{h \rightarrow 0} \frac{g(x) f(x+h) - f(x) g(x+h)}{h \cdot g(x+h) g(x)}}\\\\ & = & \displaystyle{\lim_{h \rightarrow 0} \frac{g(x) f(x+h) {\color{red}{\, - g(x) f(x)}} - f(x) g(x+h) {\color{red}{\, + g(x) f(x)}}}{h \cdot g(x+h) g(x)}} \end{array}$$We then manipulate this expression using a bit of algebra and the limit laws to reveal the same expressions that appear in the limit definitions of $f'(x)$ and $g'(x)$.
$$\begin{array}{rcl} \dfrac{d}{dx} \left[ \dfrac{f(x)}{g(x)} \right] & = & \displaystyle{\lim_{h \rightarrow 0} \dfrac{ g(x) \left[ \dfrac{f(x+h) - f(x)}{h} \right] - f(x) \left[ \dfrac{g(x+h) - g(x)}{h} \right]} {g(x+h)g(x)}}\\\\ & = & \displaystyle{\dfrac{ \displaystyle{\lim_{h \rightarrow 0}} \left( g(x) \left[ \dfrac{f(x+h) - f(x)}{h} \right] - f(x) \left[ \dfrac{g(x+h) - g(x)}{h} \right] \right)} {\displaystyle{\lim_{h \rightarrow 0} \, [g(x+h)g(x)]}}}\\\\ & = & \displaystyle{ \dfrac{ \left[ {\color{green}{ \displaystyle{ \lim_{h \rightarrow 0}} \, g(x)}} \right] \left[ {\color{red}{ \displaystyle{ \lim_{h \rightarrow 0}} \dfrac{f(x+h) - f(x)}{h} }} \right] - \left[ {\color{green}{ \displaystyle{ \lim_{h \rightarrow 0}} \, f(x)}} \right] \left[ {\color{red}{ \displaystyle{ \lim_{h \rightarrow 0}} \dfrac{g(x+h) - g(x)}{h} }} \right] } { \displaystyle{ \left[ {\color{blue}{\lim_{h \rightarrow 0} \, g(x+h)}} \right] \cdot \left[ {\color{green}{ \lim_{h \rightarrow 0} \, g(x)}}\right] }}}\\\\ \end{array}$$The limits in red are those we sought to reveal -- the limit definitions of $f'(x)$ and $g'(x)$. As for the remaining limits, note that: 1) the limits in green are easy to evaluate as their associated expressions don't depend on $h$; and 2) to evaluate the limit in blue, we must appeal to the fact that if $g$ is differentiable at $x$, it must also be continuous there -- and as such:
$$\lim_{h \rightarrow 0} g(x+h) = \lim_{u \rightarrow x} g(u) = g(x)$$Evaluating each limit, we finally arrive at the "quotient rule":
$$\dfrac{d}{dx} \left[ \dfrac{f(x)}{g(x)} \right] = \dfrac{g(x) f'(x) - f(x) g'(x)}{[g(x)]^2}$$ QED.