### Quotient Rule

Let $y=\frac{f\left(x\right)}{g\left(x\right)}$ , then

$\frac{dy}{dx}=\underset{h\to 0}{\mathrm{lim}}\frac{\frac{f\left(x+h\right)}{g\left(x+h\right)}-\frac{f\left(x\right)}{g\left(x\right)}}{h}$

$=\underset{h\to 0}{\mathrm{lim}}\frac{\frac{f\left(x+h\right)g\left(x\right)-f\left(x\right)g\left(x+h\right)}{g\left(x+h\right)g\left(x\right)}}{h}$

$=\underset{h\to 0}{\mathrm{lim}}\frac{f\left(x+h\right)g\left(x\right)-f\left(x\right)g\left(x+h\right)}{hg\left(x\right)g\left(x+h\right)}$

$=\underset{h\to 0}{\mathrm{lim}}\frac{f\left(x+h\right)g\left(x\right)-f\left(x\right)g\left(x\right)-f\left(x\right)g\left(x+h\right)+f\left(x\right)g\left(x\right)}{hg\left(x\right)g\left(x+h\right)}$

$=\underset{h\to 0}{\mathrm{lim}}\frac{f\left(x+h\right)g\left(x\right)-f\left(x\right)g\left(x\right)}{hg\left(x\right)g\left(x+h\right)}-\frac{f\left(x\right)g\left(x+h\right)-f\left(x\right)g\left(x\right)}{hg\left(x\right)g\left(x+h\right)}$

$=\underset{h\to 0}{\mathrm{lim}}\frac{\left[f\left(x+h\right)-f\left(x\right)\right]g\left(x\right)}{hg\left(x\right)g\left(x+h\right)}-\frac{f\left(x\right)\left[g\left(x+h\right)-g\left(x\right)\right]}{hg\left(x\right)g\left(x+h\right)}$

$=\underset{h\to 0}{\mathrm{lim}}\frac{f\left(x+h\right)-f\left(x\right)}{h}\frac{g\left(x\right)}{g\left(x\right)g\left(x+h\right)}-\underset{h\to 0}{\mathrm{lim}}\frac{f\left(x\right)}{g\left(x\right)g\left(x+h\right)}\frac{g\left(x+h\right)-g\left(x\right)}{h}$

$=\underset{h\to 0}{\mathrm{lim}}\frac{f\left(x+h\right)-f\left(x\right)}{h}\underset{h\to 0}{\mathrm{lim}}\frac{g\left(x\right)}{g\left(x\right)g\left(x+h\right)}-\underset{h\to 0}{\mathrm{lim}}\frac{f\left(x\right)}{g\left(x\right)g\left(x+h\right)}\underset{h\to 0}{\mathrm{lim}}\frac{g\left(x+h\right)-g\left(x\right)}{h}$

$={f}^{\prime }\left(x\right)\cdot \frac{g\left(x\right)}{{\left[g\left(x\right)\right]}^{2}}-\frac{f\left(x\right)}{{\left[g\left(x\right)\right]}^{2}}\cdot {g}^{\prime }\left(x\right)$

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