proof of quotient rule (using product rule)

Suppose $f$ and $g$ are differentiable functions defined on some intervalof $ℝ$, and $g$ never vanishes. Let us prove that

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

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

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

$=$$f^{\prime }{g}^{-1}+f{(g^{-1})}^{\prime }$

$=$$f^{\prime }{g}^{-1}+f(-1)g^{-2}{g}^{\prime }$

$=$${\displaystyle \frac{f^{\prime }}{g}}-{\displaystyle \frac{f{g}^{\prime }}{g^2}}$

$=$${\displaystyle \frac{f^{\prime }g-f{g}^{\prime }}{g^2}}.$

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