## Calculus: Early Transcendentals 8th Edition

1) First, write $\frac{f(x)}{g(x)}$ into $f(x)[g(x)]^{-1}$ 2) Then, find its derivative normally using the Product Rule. 3) Use the Chain Rule to find $[[g(x)]^{-1}]'$ 4) Simplify and modify to get the Quotient Rule.
Consider $\frac{f(x)}{g(x)}$ $$\frac{f(x)}{g(x)}=f(x)[g(x)]^{-1}$$ So, its derivative would be $$[\frac{f(x)}{g(x)}]'=[f(x)[g(x)]^{-1}]'$$ $$[\frac{f(x)}{g(x)}]'=f'(x)[g(x)]^{-1}+f(x)[[g(x)]^{-1}]'$$ *Now consider $[[g(x)]^{-1}]'$ $$[[g(x)]^{-1}]'=\frac{d[[g(x)]^{-1}]}{dx}$$ Apply the Chain Rule, $$[[g(x)]^{-1}]'=\frac{d[[g(x)]^{-1}]}{dg(x)}\frac{dg(x)}{dx}$$ $$[[g(x)]^{-1}]'=-[g(x)]^{-2}g'(x)$$ Therefore, $$[\frac{f(x)}{g(x)}]'=f'(x)[g(x)]^{-1}+f(x)[-[g(x)]^{-2}g'(x)]$$ $$[\frac{f(x)}{g(x)}]'=\frac{f'(x)}{g(x)}-\frac{f(x)g'(x)}{[g(x)]^2}$$ $$[\frac{f(x)}{g(x)}]'=\frac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^2}$$ That is the formula of the Quotient Rule. It has been proven using the Chain Rule.