Quotient rule

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}} In calculus, the quotient rule is a method of finding the derivative of a function that is the ratio of two differentiable functions. Let $h(x)=\frac{f(x)}{g(x)}$, where both Template:Mvar and Template:Mvar are differentiable and $g(x)\ne 0.$ The quotient rule states that the derivative of Template:Math is

$h^{\prime }(x)=\frac{f^{\prime }(x)g(x)-f(x)g^{\prime }(x)}{(g(x){)}^2}.$

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

Examples

Example 1: Basic example

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

Example 2: Derivative of tangent function

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

Reciprocal rule

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

Utilizing the chain rule yields the same result.

Proofs

Proof from derivative definition and limit properties

Let $h(x)=\frac{f(x)}{g(x)}.$ 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:$$\begin{array}{cc}{h}^{\prime }(x)& =\underset{k\to 0}{\lim }\frac{h(x+k)-h(x)}{k}\\ & =\underset{k\to 0}{\lim }\frac{\frac{f(x+k)}{g(x+k)}-\frac{f(x)}{g(x)}}{k}\\ & =\underset{k\to 0}{\lim }\frac{f(x+k)g(x)-f(x)g(x+k)}{k\cdot g(x)g(x+k)}\\ & =\underset{k\to 0}{\lim }\frac{f(x+k)g(x)-f(x)g(x+k)}{k}\cdot \underset{k\to 0}{\lim }\frac{1}{g(x)g(x+k)}\\ & =\underset{k\to 0}{\lim }\left[\frac{f(x+k)g(x)-f(x)g(x)+f(x)g(x)-f(x)g(x+k)}{k}\right]\cdot \frac{1}{[g(x){]}^2}\\ & =[\underset{k\to 0}{\lim }\frac{f(x+k)g(x)-f(x)g(x)}{k}-\underset{k\to 0}{\lim }\frac{f(x)g(x+k)-f(x)g(x)}{k}]\cdot \frac{1}{[g(x){]}^2}\\ & =[\underset{k\to 0}{\lim }\frac{f(x+k)-f(x)}{k}\cdot g(x)-f(x)\cdot \underset{k\to 0}{\lim }\frac{g(x+k)-g(x)}{k}]\cdot \frac{1}{[g(x){]}^2}\\ & =\frac{f^{\prime }(x)g(x)-f(x)g^{\prime }(x)}{[g(x){]}^2}.\end{array}$$The limit evaluation $\underset{k\to 0}{\lim }\frac{1}{g(x+k)g(x)}=\frac{1}{[g(x){]}^2}$ is justified by the differentiability of $g(x)$, implying continuity, which can be expressed as $\underset{k\to 0}{\lim }g(x+k)=g(x)$.

Proof using implicit differentiation

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

The product rule then gives $f^{\prime }(x)=g^{\prime }(x)h(x)+g(x)h^{\prime }(x).$

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

Proof using the reciprocal rule or chain rule

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

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

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

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

Proof by logarithmic differentiation

Let $h(x)=\frac{f(x)}{g(x)}.$ 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, $$\frac{h^{\prime }(x)}{h(x)}=\frac{f^{\prime }(x)}{f(x)}-\frac{g^{\prime }(x)}{g(x)}$$

Solving for $h^{\prime }(x)$ and substituting back $\frac{f(x)}{g(x)}$ for $h(x)$ gives: $$\begin{array}{cc}{h}^{\prime }(x)& =h(x)[\frac{f^{\prime }(x)}{f(x)}-\frac{g^{\prime }(x)}{g(x)}]\\ & =\frac{f(x)}{g(x)}[\frac{f^{\prime }(x)}{f(x)}-\frac{g^{\prime }(x)}{g(x)}]\\ & =\frac{f^{\prime }(x)}{g(x)}-\frac{f(x)g^{\prime }(x)}{g(x{)}^2}\\ & =\frac{f^{\prime }(x)g(x)-f(x)g^{\prime }(x)}{g(x{)}^2}.\end{array}$$

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 $\frac{d}{dx}(\ln |u|)=\frac{u^{\prime }}{u}$, which justifies taking the absolute value of the functions for logarithmic differentiation.

Higher order derivatives

Implicit differentiation can be used to compute the Template:Mvarth derivative of a quotient (partially in terms of its first Template:Math derivatives). For example, differentiating $f=gh$ twice (resulting in $f^{″}=g^{″}h+2g^{\prime }{h}^{\prime }+g{h}^{″}$) and then solving for $h^{″}$ yields$$h^{″}=\left(\frac{f}{g}\right)=\frac{f^{″}-g^{″}h-2g^{\prime }{h}^{\prime }}{g}.$$

See also

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