Quotient Rule

The quotient rule is useful for finding the derivatives of rational functions.

Quotient Rule

In Lagrange's notation,

$\left(\frac{p}{q}\right)' = \frac{q \cdot p' - p\cdot q'}{q^2}$

given that $\displaystyle q \neq 0$ and $\displaystyle p'$ and $q'$ exist.

Let's take a look at this in action.

Find the derivative of

$f(x)=\frac{x^2+1}{x}.$

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Since $(x^2+1)' = 2x$ and $(x)' = 1,$

$\begin{aligned} f'(x) &=\frac{x(2x)-(1)\big(x^2+1\big)}{x^2}\\ &=\frac{2x^2-x^2-1}{x^2}\\ &=\frac{x^2-1}{x^2}. \ _\square \end{aligned}$

Find the derivative of

$f(x) = \frac{e^x}{x^2}.$

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Since $(e^x)'=e^x$ and $\big(x^2\big)'=2x,$

$\begin{aligned} f'(x) &=\frac{\big(x^2\big)(e^x)-(2x)(e^x)}{x^4}\\ &= \frac{x^2e^x-2xe^x}{x^4}\\ &=\frac{xe^x-2e^x}{x^3}.\ _ \square \end{aligned}$

If $\displaystyle f(x)=\frac{\sin x}{x^3} ,$ what is $f'(x)?$

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Since $(\sin x)'=\cos x$ and $\left(x^3\right)'=3x^2,$

$\begin{aligned} f'(x)&=\frac{\left(x^3\right)\left( \cos x \right) - \left(\sin x\right)\left(3x^2\right)}{x^6} \\ &= \frac{x^2\left(x \cos x - 3 \sin x\right)}{x^6} \\ &= \frac{x \cos x - 3 \sin x}{x^4}. \ _\square \end{aligned}$

Some problems call for the combined use of differentiation rules:

Find the derivative of

$f(x) = \frac{e^{\cos x}+\tan x}{e^{3x}}.$

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Since $(e^{\cos x} + \tan x)'=-\sin x e^{\cos x} + \sec ^2 x$ and $\big(e^{3x}\big)' = 3e^{3x},$

$\begin{aligned} f'(x) &=\frac{\big(e^{3x}\big)(-\sin x e^{\cos x} + \sec ^2 x)-\big(3e^{3x}\big)(e^{\cos x} + \tan x)}{(e^{3x})^2}\\ &=\frac{(-\sin x e^{\cos x} + \sec ^2 x)-3(e^{\cos x} + \tan x)}{e^{3x}}\\ &=\frac{-\sin x e^{\cos x} + \sec ^2 x -3e^{\cos x} - 3\tan x}{e^{3x}}. \ _\square \end{aligned}$

If that last example was confusing, visit the page on the chain rule.

Now it's time to look at the proof of the quotient rule:

Let $h(x)=\frac{f(x)}{g(x)}$ and assume that $\displaystyle g(x) \neq 0$ and $f'(x)$ and $g'(x)$ exist.

Using the limit definition of the derivative,

$h'(x) = \left(\frac{f(x)}{g(x)}\right)' = \lim_{h\to 0} \frac{\frac{f(x+h)}{g(x+h)}-\frac{f(x)}{g(x)}}{h} = \lim_{h\to 0} \frac{1}{h} \frac{f(x+h)g(x)-f(x)g(x+h)}{g(x+h)g(x)}.$

Note that adding $\displaystyle -f(x)g(x) + f(x)g(x)$ to the numerator does not change the value of the fraction:

$\begin{aligned} \lim_{h\to 0} \frac{1}{h} \frac{f(x+h)g(x)-f(x)g(x+h)}{g(x+h)g(x)} &= \lim_{h\to 0} \frac{1}{h} \frac{f(x+h)g(x)-f(x)g(x) + f(x)g(x)-f(x)g(x+h)}{g(x+h)g(x)}\\\\ &=\lim_{h\to 0} \frac{1}{h} \left(\frac{f(x+h)g(x)-f(x)g(x)}{g(x+h)g(x)} - \frac{f(x)g(x+h)-f(x)g(x)}{g(x+h)g(x)}\right)\\\\ &=\lim_{h\to 0}\left(\frac{1}{g(x+h)}\right) \lim_{h\to 0} \left(\frac{f(x+h)-f(x)}{h}\right) - \lim_{h\to 0} \left(\frac{g(x+h)-g(x)}{h}\right) \lim_{h\to 0} \left(\frac{f(x)}{g(x+h)g(x)}\right)\\\\ &=\frac{f'(x)}{g(x)} - \frac{f(x)g'(x)}{\big(g(x)\big)^2} \\\\ &= \frac{g(x)f'(x) - f(x)g'(x)}{\big(g(x)\big)^2}. \end{aligned}$

Letting $\displaystyle p = f(x)$ and $\displaystyle q = g(x)$ will give the resulting form at the top of the page. $_\square$