#### Video introduction

*The quotient rule for differentiation.*

#### Summary

The quotient rule for differentiation follows directly from the product rule with just a few manipulations. Let $f$ be the quotient of $g$ and $h$, \begin{align*} f(x) = \frac{g(x)}{h(x)}, \end{align*} where $h(x) \ne 0$. We want a formula for $f'(x)$ in terms of $g$, $h$, and their derivatives. We can find a relationship among $f$, $g$, $h$, and their derivatives by multiplying through by $h$ \begin{align*} f(x)h(x) = g(x) \end{align*} and then applying the product rule \begin{align*} f'(x)h(x) + f(x)h'(x) = g'(x) \end{align*} to get the derivative of $g$ in terms of $f$, $h$, and their derivatives. We subtract $f(x)h'(x)$ and divide by $h(x)$ to solve for $f'(x)$, \begin{align*} f'(x)= \frac{g'(x) -f(x)h'(x)}{h(x)}. \end{align*}

We are getting closer, but we want a formula for $f'$ just in terms of $g$, $h$, and their derivatives. We can get rid of the $f(x)$ on the right hand side by substituting the definition $f(x)=g(x)/h(x)$, \begin{align*} f'(x)= \frac{g'(x) -\frac{g(x)}{h(x)}h'(x)}{h(x)}, \end{align*} and eliminate the fraction in the numerator by multiplying numerator and denominator by $h(x)$, \begin{align*} f'(x)= \frac{h(x)g'(x) -g(x)h'(x)}{[h(x)]^2}. \end{align*} We have derived the quotient rule for the differentiation. We could rewrite it explicitly as the derivative of a quotient, \begin{align} \left(\frac{g(x)}{h(x)}\right)'= \frac{h(x)g'(x) -g(x)h'(x)}{[h(x)]^2} \label{quotient_rule1} \end{align} or \begin{align} \diff{}{x}\left(\frac{g}{h}\right)= \frac{h\diff{g}{x} -g\diff{h}{x}}{h^2}. \label{quotient_rule2} \end{align}