Quotient rule

Template:Short description Template:Calculus In calculus, the quotient rule is a method of finding the derivative of a function that is the ratio of two differentiable functions.<ref>Template:Cite book</ref><ref>Template:Cite book</ref><ref>Template:Cite book</ref> Let <math>h(x)=\frac{f(x)}{g(x)}</math>, where both Template:Mvar and Template:Mvar are differentiable and <math>g(x)\neq 0.</math> The quotient rule states that the derivative of Template:Math is

<math>h'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2}.</math>

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

Examples

Example 1: Basic example

Given <math>h(x)=\frac{e^x}{x^2}</math>, let <math>f(x)=e^x, g(x)=x^2</math>, then using the quotient rule:<math display="block">\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}</math>

Example 2: Derivative of tangent function

The quotient rule can be used to find the derivative of <math>\tan x = \frac{\sin x}{\cos x}</math> as follows: <math display="block">\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}</math>

Reciprocal rule

Template:Main The reciprocal rule is a special case of the quotient rule in which the numerator <math>f(x)=1</math>. Applying the quotient rule gives<math display="block">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}.</math>

Utilizing the chain rule yields the same result.

Proofs

Proof from derivative definition and limit properties

Let <math>h(x) = \frac{f(x)}{g(x)}.</math> Applying the definition of the derivative and properties of limits gives the following proof, with the term <math>f(x) g(x)</math> added and subtracted to allow splitting and factoring in subsequent steps without affecting the value:<math display="block">\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}</math>The limit evaluation <math>\lim_{k \to 0}\frac{1}{g(x+k)g(x)}=\frac{1}{[g(x)]^2}</math> is justified by the differentiability of <math>g(x)</math>, implying continuity, which can be expressed as <math>\lim_{k \to 0}g(x+k) = g(x)</math>.

Proof using implicit differentiation

Let <math>h(x) = \frac{f(x)}{g(x)},</math> so that <math>f(x) = g(x)h(x).</math>

The product rule then gives <math>f'(x)=g'(x)h(x) + g(x)h'(x).</math>

Solving for <math>h'(x)</math> and substituting back for <math>h(x)</math> gives: <math display="block">\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}</math>

Proof using the reciprocal rule or chain rule

Let <math>h(x) = \frac{f(x)}{g(x)} = f(x) \cdot \frac{1}{g(x)}.</math>

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

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

Substituting the result into the expression gives<math display="block">\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}</math>

Proof by logarithmic differentiation

Let <math>h(x)=\frac{f(x)}{g(x)}.</math> Taking the absolute value and natural logarithm of both sides of the equation gives <math display="block">\ln|h(x)|=\ln\left|\frac{f(x)}{g(x)}\right|</math>

Applying properties of the absolute value and logarithms, <math display="block">\ln|h(x)|=\ln|f(x)|-\ln|g(x)|</math>

Taking the logarithmic derivative of both sides, <math display="block">\frac{h'(x)}{h(x)}=\frac{f'(x)}{f(x)}-\frac{g'(x)}{g(x)}</math>

Solving for <math>h'(x)</math> and substituting back <math>\tfrac{f(x)}{g(x)}</math> for <math>h(x)</math> gives: <math display="block">\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}</math>

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 <math>\tfrac{d}{dx}(\ln|u|)=\tfrac{u'}{u}</math>, 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 <math>f=gh</math> twice (resulting in <math>f = gh + 2g'h' + gh</math>) and then solving for <math>h</math> yields<math display="block">h = \left(\frac{f}{g}\right) = \frac{f-gh-2g'h'}{g}.</math>

See also

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