Answer
\[\left(\frac{f(x)}{g(x)}\right)'=\frac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^2}\]
Work Step by Step
We can rewrite \[\frac{f(x)}{g(x)}=f(x)\cdot [g(x)]^{-1}\]
Differentiate with respect to $x$ using chain rule
\[\left(\frac{f(x)}{g(x)}\right)'=f(x)\cdot \{[g(x)]^{-1}\}'+\{f(x)\}'[g(x)]^{-1}\]
\[\left(\frac{f(x)}{g(x)}\right)'=(-1)f(x)\cdot [g(x)]^{-2}+f'(x)[g(x)]^{-1}\]
\[\left(\frac{f(x)}{g(x)}\right)'=-\frac{f(x)}{[g(x)]^2}+\frac{f'(x)}{[g(x)]}\]
\[\left(\frac{f(x)}{g(x)}\right)'=\frac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^2}\]
Hence \[\left(\frac{f(x)}{g(x)}\right)'=\frac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^2}\]
