proof of quotient rule

say we have an expression $\frac{m(x)}{n(x)}$ and we need to find the derivative.

the quotient rule says that $\frac{d}{dx}[\frac{m(x)}{n(x)}]=\frac{m^{\prime }(x)n(x)-m(x)n^{\prime }(x)}{n(x{)}^2}$

let’s do it step by step

we solve for $\frac{d}{dx}[m(x)\cdot n(x{)}^{-1}]$

product rule

$\frac{d}{dx}[m(x)\cdot n(x{)}^{-1}]=m^{\prime }(x)(n(x{)}^{-1})+m(x)\frac{d}{dx}[n(x{)}^{-1}]$

chain rule

$\frac{d}{dx}[n(x{)}^{-1}]=(-1)(n(x){)}^{-2}\cdot n^{\prime }(x)$

$\frac{d}{dx}[n(x{)}^{-1}]=-n(x{)}^{-2}\cdot n^{\prime }(x)$

$\frac{d}{dx}[m(x)\cdot n(x{)}^{-1}]=m^{\prime }(x)(n(x{)}^{-1})+m(x)[-n(x{)}^{-2}\cdot n^{\prime }(x)]$

$\frac{d}{dx}[m(x)\cdot n(x{)}^{-1}]=\frac{m^{\prime }(x)}{n(x)}+\frac{-m(x)n^{\prime }(x)}{n(x{)}^2}$

$\frac{d}{dx}[m(x)\cdot n(x{)}^{-1}]=\frac{m^{\prime }(x)n(x)}{n(x{)}^2}+\frac{-m(x)n^{\prime }(x)}{n(x{)}^2}$

$\frac{d}{dx}[m(x)\cdot n(x{)}^{-1}]=\frac{m^{\prime }(x)n(x)-m(x)n^{\prime }(x)}{n(x{)}^2}$

interesting, we did it!