Quotient Rule

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Quotient Rule for Derivatives – HMC Calculus Tutorial

Suppose we are working with a function $h(x)$ that is a ratio of two functions $f(x)$ and $g(x)$.

How is the derivative of $h(x)$ related to $f(x)$, $g(x)$, and their derivatives?

Quotient Rule

Let $f$ and $g$ be differentiable at $x$ with $g(x)\neq 0$. Then $f/g$ is differentiable at $x$ and \[\left[\frac{f(x)}{g(x)}\right]’=\frac{g(x)f'(x)-f(x)g'(x)}{[g(x)]^2}.\]

We will apply the limit definition of the derivative:

\[f'(x)=\lim_{\Delta x\to 0}\, \frac{f(x+\Delta x)-f(x)}{\Delta x}\]

\begin{eqnarray*} h'(x)=\left[\frac{f(x)}{g(x)}\right]’&=&\lim_{h\to 0}\, \frac{\frac{f(x+h)}{g(x+h)}-\frac{f(x)}{g(x)}}{h}\\ &=& \lim_{h\to 0}\, \frac{1}{h}\, \frac{f(x+h)g(x)-f(x)g(x+h)}{g(x+h)g(x)}\\ &=& \lim_{h\to 0}\, \frac{1}{h}\, \frac{f(x+h)g(x)-f(x)g(x)+f(x)g(x)-f(x)g(x+h)}{g(x+h)g(x)}\\ &=& \lim_{h\to 0}\, \frac{1}{h}\left[ \frac{f(x+h)g(x)-f(x)g(x)}{g(x+h)g(x)}-\frac{f(x)g(x+h)-f(x)g(x)}{g(x+h)g(x)} \right]\\ &=& \lim_{h\to 0}\left[\frac{1}{g(x+h)}\right]\, \lim_{h\to 0}\left[\frac{f(x+h)-f(x)}{h}\right]\\ &~&\qquad -\lim_{h\to 0} \left[\frac{f(x)}{g(x+h)g(x)}\right]\, \lim_{h\to 0} \left[\frac{g(x+h)-g(x)}{h}\right]\\ &=& \frac{1}{g(x)}\cdot f'(x)-\frac{f(x)}{[g(x)]^2}\cdot g'(x)\\ &=&\frac{g(x)f'(x)-f(x)g'(x)}{[g(x)]^2}. \end{eqnarray*} We have implicitly assumed here that $f'(x)$ and $g'(x)$ exist and that $g(x)\neq 0$:

\begin{eqnarray*} h'(x)=\left[\frac{f(x)}{g(x)}\right]’&=&\lim_{h\to 0}\, \frac{\frac{f(x+h)}{g(x+h)}-\frac{f(x)}{g(x)}}{h}\\ &=& \lim_{h\to 0}\, \frac{1}{h}\, \frac{f(x+h)g(x)-f(x)g(x+h)}{g(x+h)g(x)}\\ &=& \lim_{h\to 0}\, \frac{1}{h}\, \frac{f(x+h)g(x)-f(x)g(x)+f(x)g(x)-f(x)g(x+h)}{g(x+h)g(x)}\\ &=& \lim_{h\to 0}\, \frac{1}{h}\left[ \frac{f(x+h)g(x)-f(x)g(x)}{g(x+h)g(x)}-\frac{f(x)g(x+h)-f(x)g(x)}{g(x+h)g(x)} \right]\\ &=& \lim_{h\to 0}\left[\frac{1}{g(x+h)}\right]\, \lim_{h\to 0}\left[\frac{f(x+h)-f(x)}{h}\right]\\ &~&\qquad -\lim_{h\to 0} \left[\frac{f(x)}{g(x+h)g(x)}\right]\, \lim_{h\to 0} \left[\frac{g(x+h)-g(x)}{h}\right]\\ &=& \frac{1}{g(x)}\cdot f'(x)-\frac{f(x)}{[g(x)]^2}\cdot g'(x)\\ &=&\frac{g(x)f'(x)-f(x)g'(x)}{[g(x)]^2}. \end{eqnarray*} We have implicitly assumed here that $f'(x)$ and $g'(x)$ exist and that $g(x)\neq 0$.

Examples
  • If $\displaystyle f(x)=\frac{2x+1}{x-3}$, then \begin{eqnarray*} f'(x)&=&\frac{(x-3)\frac{d}{dx}[2x+1]-(2x+1)\frac{d}{dx}[x-3]}{[x-3]^2}\\ &=& \frac{(x-3)(2)-(2x+1)(1)}{(x-3)^2}\\ &=& -\frac{7}{(x-3)^2}. \end{eqnarray*}

  • If $\displaystyle f(x)=\tan x=\frac{\sin x}{\cos x}$, then \begin{eqnarray*} f'(x)&=&\frac{\cos (x)\frac{d}{dx}[\sin (x)]-\sin (x)\frac{d}{dx} [\cos x]}{[\cos x ]^2}\\ &=& \frac{\cos^2 (x)+\sin^2 (x)}{\cos^2 (x)}\\ &=& \frac{1}{\cos^2 (x)}\\ &=& \sec^2 (x), \end{eqnarray*} verifying the familiar differentiation formula for $\tan (x)$.

  • If $\displaystyle f(x)=\frac{1}{g(x)}$, then \begin{eqnarray*} f'(x)=\left[\frac{1}{g(x)}\right]’&=&\frac{g(x)\frac{d}{dx}[1]-(1)g'(x)} {[g(x)]^2}\\ &=& \frac{g(x)(0)-(1)g'(x)}{[g(x)]^2}\\ &=& -\frac{g'(x)}{[g(x)]^2}. \end{eqnarray*} For example, $\displaystyle \frac{d}{dx}[x^{-4}]=\frac{d}{dx}\left[\frac{1}{x^4}\right] =-\frac{\frac{d}{dx}[x^4]}{[x^4]^2}=-\frac{4x^3}{x^8}=-\frac{4}{x^5}=-4x^{-5}$.

Key Concepts

Quotient Rule

Let $f$ and $g$ be differentiable at $x$ with $g(x) \neq 0$. Then $f/g$ is differentiable at $x$ and $$\left[\frac{f(x)}{g(x)}\right]’ = \frac{g(x)f'(x)-f(x)g'(x)}{\left[g(x)\right]^2}.$$

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