Quotient rule

In calculus, the quotient rule is a method of finding the derivative of a function that is the ratio of two differentiable functions. Let ⁠\(\textstyle h(x) = \frac{f(x)}{g(x)}\)⁠, where both ⁠\(f\)⁠ and ⁠\(g\)⁠ are differentiable and ⁠\(g(x)\neq 0\)⁠. The quotient rule states that the derivative of ⁠\(h(x)\)⁠ is \[h'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2}.\]

It is provable in many ways by using other derivative rules.

Given ⁠\(\textstyle h(x) = \frac{e^x}{x^2}\)⁠, let ⁠\(f(x) = e^x\)⁠, ⁠\(g(x) = x^2\)⁠, then using the quotient rule: \[\begin{aligned} \frac{d}{dx} \left(\frac{e^x}{x^2}\right) &= \frac{\left(\frac{d}{dx}e^x\right)(x^2) - (e^x)\left(\frac{d}{dx} x^2\right)}{(x^2)^2} \\ &= \frac{(e^x)(x^2) - (e^x)(2x)}{x^4} \\ &= \frac{x^2 e^x - 2x e^x}{x^4} \\ &= \frac{x e^x - 2 e^x}{x^3} \\ &= \frac{e^x(x - 2)}{x^3}. \end{aligned}\]

The quotient rule can be used to find the derivative of \(\tan x = \frac{\sin x}{\cos x}\) as follows: \[\begin{aligned} \frac{d}{dx} \tan x &= \frac{d}{dx} \left(\frac{\sin x}{\cos x}\right) \\ &= \frac{\left(\frac{d}{dx}\sin x\right)(\cos x) - (\sin x)\left(\frac{d}{dx}\cos x\right)}{\cos^2 x} \\ &= \frac{(\cos x)(\cos x) - (\sin x)(-\sin x)}{\cos^2 x} \\ &= \frac{\cos^2 x + \sin^2 x}{\cos^2 x} \\ &= \frac{1}{\cos^2 x} = \sec^2 x. \end{aligned}\]

The reciprocal rule is a special case of the quotient rule in which the numerator ⁠\(f(x)=1\)⁠. Applying the quotient rule gives \[h'(x)=\frac{d}{dx}\left[\frac{1}{g(x)}\right]=\frac{0 \cdot g(x) - 1 \cdot g'(x)}{g(x)^2}=\frac{-g'(x)}{g(x)^2}.\]

Utilizing the chain rule yields the same result.

Let ⁠\(\textstyle h(x) = \frac{f(x)}{g(x)}\)⁠. Applying the definition of the derivative and properties of limits gives the following proof, with the term \(f(x) g(x)\) added and subtracted to allow splitting and factoring in subsequent steps without affecting the value: \[\begin{aligned} h'(x) &= \lim_{k\to 0} \frac{h(x+k) - h(x)}{k} \\ &= \lim_{k\to 0} \frac{\frac{f(x+k)}{g(x+k)} - \frac{f(x)}{g(x)}}{k} \\ &= \lim_{k\to 0} \frac{f(x+k)g(x) - f(x)g(x+k)}{k \cdot g(x)g(x+k)} \\ &= \lim_{k\to 0} \frac{f(x+k)g(x) - f(x)g(x+k)}{k} \cdot \lim_{k\to 0}\frac{1}{g(x)g(x+k)} \\ &= \lim_{k\to 0} \left[\frac{f(x+k)g(x) - f(x)g(x) + f(x)g(x) - f(x)g(x+k)}{k} \right] \cdot \frac{1}{[g(x)]^2} \\ &= \left[\lim_{k\to 0} \frac{f(x+k)g(x) - f(x)g(x)}{k} - \lim_{k\to 0}\frac{f(x)g(x+k) - f(x)g(x)}{k} \right] \cdot \frac{1}{[g(x)]^2} \\ &= \left[\lim_{k\to 0} \frac{f(x+k) - f(x)}{k} \cdot g(x) - f(x) \cdot \lim_{k\to 0}\frac{g(x+k) - g(x)}{k} \right] \cdot \frac{1}{[g(x)]^2} \\ &= \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}. \end{aligned}\] The limit evaluation \(\lim_{k \to 0}\frac{1}{g(x+k)g(x)}=\frac{1}{[g(x)]^2}\) is justified by the differentiability of ⁠\(g(x)\)⁠, implying continuity, which can be expressed as ⁠\(\textstyle \lim_{k \to 0}g(x+k) = g(x)\)⁠.

Let ⁠\(\textstyle h(x) = \frac{f(x)}{g(x)}\)⁠, so that ⁠\(f(x) = g(x)h(x)\)⁠.

The product rule then gives ⁠\(f'(x) = g'(x)h(x) + g(x)h'(x)\)⁠.

Solving for \(h'(x)\) and substituting back for \(h(x)\) gives: \[\begin{aligned} h'(x) &= \frac{f'(x) -g'(x)h(x)}{g(x)} \\ &= \frac{f'(x) - g'(x)\cdot\frac{f(x)}{g(x)}}{g(x)} \\ &= \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}. \end{aligned}\]

Let ⁠\(\textstyle h(x) = \frac{f(x)}{g(x)} = f(x) \cdot \frac{1}{g(x)}\)⁠.

Then the product rule gives ⁠\(\textstyle h'(x) = f'(x)\cdot\frac{1}{g(x)} + f(x) \cdot \frac{d}{dx}\left[\frac{1}{g(x)}\right]\)⁠.

To evaluate the derivative in the second term, apply the reciprocal rule, or the power rule along with the chain rule: \[\begin{aligned} \frac{d}{dx}\left[\frac{1}{g(x)}\right] &= -\frac{1}{g(x)^2} \cdot g'(x) \\ &= \frac{-g'(x)}{g(x)^2} . \end{aligned}\]

Substituting the result into the expression gives \[\begin{aligned} h'(x) &= f'(x)\cdot\frac{1}{g(x)} + f(x)\cdot\left[\frac{-g'(x)}{g(x)^2}\right] \\ &= \frac{f'(x)}{g(x)} - \frac{f(x)g'(x)}{g(x)^2} \\ &= {\frac{g(x)}{g(x)}}\cdot{\frac{f'(x)}{g(x)}} - \frac{f(x)g'(x)}{g(x)^2} \\ &= \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2} . \end{aligned}\]

Let ⁠\(\textstyle h(x) = \frac{f(x)}{g(x)}\)⁠. Taking the absolute value and natural logarithm of both sides of the equation gives \[\ln|h(x)| = \ln\left|\frac{f(x)}{g(x)}\right| .\]

Applying properties of the absolute value and logarithms, \[\ln|h(x)| = \ln|f(x)| - \ln|g(x)| .\]

Taking the logarithmic derivative of both sides, \[\frac{h'(x)}{h(x)} = \frac{f'(x)}{f(x)} - \frac{g'(x)}{g(x)} .\]

Solving for \(h'(x)\) and substituting back \(\tfrac{f(x)}{g(x)}\) for \(h(x)\) gives: \[\begin{aligned} h'(x) &= h(x)\left[\frac{f'(x)}{f(x)}-\frac{g'(x)}{g(x)}\right] \\ &= \frac{f(x)}{g(x)}\left[\frac{f'(x)}{f(x)}-\frac{g'(x)}{g(x)}\right] \\ &= \frac{f'(x)}{g(x)}-\frac{f(x)g'(x)}{g(x)^2} \\ &= \frac{f'(x)g(x)-f(x)g'(x)}{g(x)^2} . \end{aligned}\]

Taking the absolute value of the functions is necessary for the logarithmic differentiation of functions that may have negative values, as logarithms are only real-valued for positive arguments. This works because ⁠\(\textstyle \frac{d}{dx}(\ln\vert u\vert) = \frac{u'}{u}\)⁠, which justifies taking the absolute value of the functions for logarithmic differentiation.

Implicit differentiation can be used to compute the nth derivative of a quotient (partially in terms of its first n−1 derivatives). For example, differentiating \(f=gh\) twice (resulting in ⁠\(f'' = g''h + 2g'h' + gh''\)⁠) and then solving for \(h''\) yields \[h'' = \left(\frac{f}{g}\right)'' = \frac{f''-g''h-2g'h'}{g}.\]