Introduction
We will prove the following (Assuming the denominator is not
0
):
\frac{d}{dx}\frac{f(x)}{g(x)}=\frac{f(x)g'(x)+f'(x)g(x)}{g(x)^2}
Proof
\begin{align*}
\frac{d}{dx}f(x)g(x)&=\lim_{h \to 0}\frac{\frac{f(x+h)}{g(x+h)} - \frac{f(x)}{g(x)}}{h} \\\\
&=\lim_{h \to 0}\frac{1}{h} \cdot \left[ \frac{f(x+h)}{g(x+h)} - \frac{f(x)}{g(x)} \right] \\\\
&=\lim_{h \to 0}\frac{1}{h} \cdot \frac{f(x+h) g(x) - f(x) g(x+h)}{g(x) g(x+h)} \\\\
&=\lim_{h \to 0}\frac{1}{h} \cdot \frac{f(x+h) g(x) \c1{- f(x)g(x) + f(x)g(x)} - f(x) g(x+h)}{g(x) g(x+h)} \\\\
&=\lim_{h \to 0}\c1{\frac{1}{h g(x) g(x+h)}} \left[f(x+h) g(x) - f(x) g(x) + f(x) g(x) - f(x) g(x+h) \right] \\\\
&=\lim_{h \to 0}\frac{1}{hg(x)g(x+h)} \c1{\left[ g(x) [f(x+h) - f(x)] + f(x) [g(x) - g(x+h)] \right]} \\\\
&=\lim_{h \to 0}\frac{1}{g(x) g(x+h)} \left[ g(x) \frac{f(x+h) - f(x)}{\c1{h}} + f(x) \frac{g(x) - g(x+h)}{\c1{h}} \right] \\\\
&=\c1{\lim_{h \to 0}}\frac{1}{g(x) g(x+h)} \left[ \c1{\lim_{h \to 0}}g(x) \c1{\lim_{h \to 0}}\frac{f(x+h) - f(x)}{h} + \c1{\lim_{h \to 0}}f(x) \c1{\lim_{h \to 0}}\frac{g(x) - g(x+h)}{h} \right] \\\\
&=\frac{1}{g(x)^2} \left[ g(x) f'(x) + f(x) g'(x) \right] \\\\
&=\frac{g(x) f'(x) + f(x) g'(x)}{\c1{g(x)^2}}
\end{align*}