Suppose $f$ and $g$ are differentiable functions defined on some interval of , and $g$ never vanishes. Let us prove that $$ \left( \frac{f}{g}\right)' = \frac{ f' g - f g'}{g^2}. $$

Using the product rule $(fg)'=f'g + fg'$ , and $(g^{-1})'=-g^{-2}g'$ , we have \begin{eqnarray*} \left( \frac{f}{g}\right)' &=& (f g^{-1})' \\ &=& f' g^{-1} + f (g^{-1})' \\ &=& f' g^{-1} + f (-1) g^{-2} g'\\ &=& \frac{f'}{g} - \frac{fg'}{g^2} \\ &=& \frac{f'g- fg'}{g^2}. \end{eqnarray*}Here $g^{-1}=1/g$ and $g^{-2}=1/g^2$ .