Here we present the proof of the following theorem:

Let f,g:\mathbb{R}\rightarrow\mathbb{R} be functions that are differentiable at some a\in\mathbb{R}.  If g(a)\neq 0, then f/g is differentiable at a with

\left(\frac{f}{g}\right)'(a)=\frac{f'(a)g(a)-f(a)g'(a)}{[g(a)]^2}

Quotient Rule

Remark: In the Leibniz notation,

\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}

Proof: Let q=f/g:

q(a+h)-q(a)=\frac{f(a+h)}{g(a+h)}-\frac{f(a)}{g(a)}

=\frac{f(a+h)g(a)-f(a)g(a+h)}{g(a+h)g(a)}

=\frac{f(a+h)g(a)\overbrace{-f(a)g(a)+f(a)g(a)}^{=0}-f(a)g(a+h)}{g(a+h)g(a)}

=\frac{g(a)[f(a+h)-f(a)]-f(a)[g(a+h)-g(a)]}{g(a+h)g(a)}

\Rightarrow \frac{q(a+h)-q(a)}{h}=\frac{g(a)\left[\frac{f(a+h)-f(a)}{h}\right]-f(a)\left[\frac{g(a+h)-g(a)}{h}\right]}{g(a+h)g(a)}

Letting h\rightarrow 0 on both sides:

q'(a)=\left(\frac{f}{g}\right)'(a)=\frac{g(a)f'(a)-f(a)g'(a)}{[g(a)]^2} \bullet

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