# proof of quotient rule (using product rule)

Suppose $f$ and $g$ are differentiable functions defined on some interval of $\mathbbmss{R}$, and $g$ never vanishes. Let us prove that

 $\left(\frac{f}{g}\right)^{\prime}=\frac{f^{\prime}g-fg^{\prime}}{g^{2}}.$

Using the product rule $(fg)^{\prime}=f^{\prime}g+fg^{\prime}$, and $(g^{-1})^{\prime}=-g^{-2}g^{\prime}$, we have

 $\displaystyle\left(\frac{f}{g}\right)^{\prime}$ $\displaystyle=$ $\displaystyle(fg^{-1})^{\prime}$ $\displaystyle=$ $\displaystyle f^{\prime}g^{-1}+f(g^{-1})^{\prime}$ $\displaystyle=$ $\displaystyle f^{\prime}g^{-1}+f(-1)g^{-2}g^{\prime}$ $\displaystyle=$ $\displaystyle\frac{f^{\prime}}{g}-\frac{fg^{\prime}}{g^{2}}$ $\displaystyle=$ $\displaystyle\frac{f^{\prime}g-fg^{\prime}}{g^{2}}.$

Here $g^{-1}=1/g$ and $g^{-2}=1/g^{2}$.

Title proof of quotient rule (using product rule) ProofOfQuotientRuleusingProductRule 2013-03-22 15:00:45 2013-03-22 15:00:45 matte (1858) matte (1858) 5 matte (1858) Proof msc 26A06