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

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:$$\begin{align} \frac{d}{dx} \left(\frac{e^x}{x^2}\right) &= \frac{\left(\frac{d}{dx}e^x\right)(x^2) - (e^x)\left(\frac{d}{dx} x^2\right)}{(x^2)^2} \\ &= \frac{(e^x)(x^2) - (e^x)(2x)}{x^4} \\ &= \frac{x^2 e^x - 2x e^x}{x^4} \\ &= \frac{x e^x - 2 e^x}{x^3} \\ &= \frac{e^x(x - 2)}{x^3}. \end{align}$$

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: $$\begin{align} \frac{d}{dx} \tan x &= \frac{d}{dx} \left(\frac{\sin x}{\cos x}\right) \\ &= \frac{\left(\frac{d}{dx}\sin x\right)(\cos x) - (\sin x)\left(\frac{d}{dx}\cos x\right)}{\cos^2 x} \\ &= \frac{(\cos x)(\cos x) - (\sin x)(-\sin x)}{\cos^2 x} \\ &= \frac{\cos^2 x + \sin^2 x}{\cos^2 x} \\ &= \frac{1}{\cos^2 x} = \sec^2 x. \end{align}$$

Reciprocal rule
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.

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:$$\begin{align} h'(x) &= \lim_{k\to 0} \frac{h(x+k) - h(x)}{k} \\ &= \lim_{k\to 0} \frac{\frac{f(x+k)}{g(x+k)} - \frac{f(x)}{g(x)}}{k} \\ &= \lim_{k\to 0} \frac{f(x+k)g(x) - f(x)g(x+k)}{k \cdot g(x)g(x+k)} \\ &= \lim_{k\to 0} \frac{f(x+k)g(x) - f(x)g(x+k)}{k} \cdot \lim_{k\to 0}\frac{1}{g(x)g(x+k)} \\ &= \lim_{k\to 0} \left[\frac{f(x+k)g(x) - f(x)g(x) + f(x)g(x) - f(x)g(x+k)}{k} \right] \cdot \frac{1}{g(x)^2} \\ &= \left[\lim_{k\to 0} \frac{f(x+k)g(x) - f(x)g(x)}{k} - \lim_{k\to 0}\frac{f(x)g(x+k) - f(x)g(x)}{k} \right] \cdot \frac{1}{g(x)^2} \\ &= \left[\lim_{k\to 0} \frac{f(x+k) - f(x)}{k} \cdot g(x) - f(x) \cdot \lim_{k\to 0}\frac{g(x+k) - g(x)}{k} \right] \cdot \frac{1}{g(x)^2} \\ &= \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2}. \end{align}$$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 $$\lim_{k \to 0}g(x+k) = g(x)$$.

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 $$f'(x)=g'(x)h(x) + g(x)h'(x).$$

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

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 $$h'(x) = f'(x)\cdot\frac{1}{g(x)} + f(x) \cdot \frac{d}{dx}\left[\frac{1}{g(x)}\right].$$

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

Substituting the result into the expression gives$$\begin{align} 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{align}$$

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, $$\frac{h'(x)}{h(x)}=\frac{f'(x)}{f(x)}-\frac{g'(x)}{g(x)}$$

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

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

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