Quotient Rule for Derivatives - HMC Calculus Tutorial

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Quotient Rule for Derivatives

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) = f(x)/g(x) related to f(x), g(x), and their derivatives?

Let f and g be differentiable at x with g(x) ≠ 0. Then f/g is differentiable at x and

$\begin{displaymath}\left[\frac{f(x)}{g(x)}\right]'=\frac{g(x)f'(x)-f(x)g'(x)}{[g(x)]^2}.\end{displaymath}$

Proof

If $\displaystyle f(x)=\frac{2x+1}{x-3}$ , then

$\begin{eqnarray*} f'(x)&=&\frac{(x-3)\frac{d}{dx}[2x+1]-(2x+1)\frac{d}{dx}[x-3]}... ...rac{(x-3)(2)-(2x+1)(1)}{(x-3)^2}\\ [6pt] &=& -\frac{7}{(x-3)^2}. \end{eqnarray*}$
If $\displaystyle f(x)=\tan x=\frac{\sin x}{\cos x}$ , then

$\begin{eqnarray*} f'(x)&=&\frac{\cos (x)\frac{d}{dx}[\sin (x)]-\sin (x)\frac{d}{... ...^2 (x)}\\ [6pt] &=& \frac{1}{\cos^2 (x)}\\ [6pt] &=& \sec^2 (x), \end{eqnarray*}$

verifying the familiar differentiation formula for tan(x).
If $\displaystyle f(x)=\frac{1}{g(x)}$ , then

$\begin{eqnarray*} f'(x)=\left[\frac{1}{g(x)}\right]'&=&\frac{g(x)\frac{d}{dx}[1]... ...g(x)(0)-(1)g'(x)}{[g(x)]^2}\\ [6pt] &=& -\frac{g'(x)}{[g(x)]^2}. \end{eqnarray*}$

For example,

$\begin{eqnarray*} \frac{d}{dx} [ \csc{x} ] &=& \left[ \frac{1}{\sin{(x)}} \right... ...^2} &=& \frac{-\cos{(x)}}{[\sin{(x)}]^2} &=& -\cot{(x)}\csc{(x)} \end{eqnarray*}$

as expected.

Key Concepts

Quotient Rule

Let f and g be differentiable at x with g(x)≠ 0. Then f/g is differentiable at x and

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