Quotient Rule

Statement

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

Proof

This proof is taken from Khan Academy.

\frac{d}{dx}\left[ \frac{f(x)}{g(x)} \right] = \frac{d}{dx}\left[ f(x)\cdot(g(x))^{-1}=f^\prime(x)(g(x))^{-1} + f(x)\left[\frac{d}{dx}(g(x))^{-1}\right]

What is the derivative of (g(x))^{-1}? It’s a little more complicated than one might think so we’ll devote a short aside to its calculation.

Let u = g(x). Then (g(x))^{-1}=u(g(x))^{-1}. To find \frac{d}{dx}u(g(x)^{-1}, we have to apply the chain rule:

\frac{du}{dx}=\frac{du}{dg}\cdot\frac{dg}{dx}=\left[ (-1)u^{-2}\right]\left[ \frac{dg}{dx}\right]=-u^{-2}g^\prime

To get this back to an equation that is all in terms of g(x), we substitute g(x) for u into the equation above. We get:

\frac{d}{dx}(g(x))^{-1}=-(g(x))^{-2}g^\prime(x)

We substitute our result for the derivative of (g(x))^{-1} back into our original expression for \frac{d}{dx}\left[ \frac{f(x)}{g(x)} \right]. We get:

\begin{array}{rcl} \frac{d}{dx}\left[ \frac{f(x)}{g(x)} \right] &=& f^\prime(x)(g(x))^{-1} + f(x)(-1)g(x)^{-2}g^\prime(x)\\ \, &\,& \, \\ &=& f^\prime(x)(g(x))^{-1}\frac{(g(x))^{-1}}{(g(x))^{-1}} + f(x)(-1)g(x)^{-2}g^\prime(x)\\ \, &\,& \, \\ &=& \frac{f^\prime(x)g(x)}{(g(x))^2} - \frac{f(x)g^\prime(x)}{(g(x))^2}\\ \, &\,& \, \\ &=& \frac{f^\prime(x)g(x) - f(x)g^\prime(x)}{(g(x))^2} \end{array}