Quotient rule

 ( Articles Containing Proofs ) 

In calculus, the quotient rule is a method of finding the derivative of a function that is the ratio of two differentiable functions.

(2025). 9780547167022, Brooks/Cole. ISBN 9780547167022

(2025). 9780321588760, Addison-Wesley. ISBN 9780321588760

Let $h(x)=\backslash frac\{f(x)\}\{g(x)\}$, where both and are differentiable and $g(x)\backslash neq\; 0.$ The quotient rule states that the derivative of is

$h\text{'}(x)\; =\; \backslash frac\{f\text{'}(x)g(x)\; -\; f(x)g\text{'}(x)\}\{(g(x))^2\}.$

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

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Examples

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Example 1: Basic example Given $h(x)=\backslash frac\{e^x\}\{x^2\}$, let $f(x)=e^x,\; g(x)=x^2$, then using the quotient rule:$$\backslash begin\{align\}$$

   \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}
     

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Example 2: Derivative of tangent function The quotient rule can be used to find the derivative of $\backslash tan\; x\; =\; \backslash frac\{\backslash sin\; x\}\{\backslash cos\; x\}$ as follows: $$\backslash begin\{align\}$$

   \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}
     

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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$$h\text{'}(x)=\backslash frac\{d\}\{dx\}\backslash left\backslash frac\{1\}\{g(x)\}\backslash right=\backslash frac\{0\; \backslash cdot\; g(x)\; -\; 1\; \backslash cdot\; g\text{'}(x)\}\{g(x)^2\}=\backslash frac\{-g\text{'}(x)\}\{g(x)^2\}.$$

Utilizing the chain rule yields the same result.

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Proofs

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Proof from derivative definition and limit properties Let $h(x)\; =\; \backslash frac\{f(x)\}\{g(x)\}.$ 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:$$\backslash begin\{align\}$$

  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 $\backslash lim\_\{k\; \backslash to\; 0\}\backslash frac\{1\}\{g(x+k)g(x)\}=\backslash frac\{1\}\{[g(x)]^2\}$ is justified by the differentiability of $g(x)$, implying continuity, which can be expressed as $\backslash lim\_\{k\; \backslash to\; 0\}g(x+k)\; =\; g(x)$.
     

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Proof using implicit differentiation Let $h(x)\; =\; \backslash frac\{f(x)\}\{g(x)\},$ so that $f(x)\; =\; g(x)h(x).$

The product rule then gives $f\text{'}(x)=g\text{'}(x)h(x)\; +\; g(x)h\text{'}(x).$

Solving for $h\text{'}(x)$ and substituting back for $h(x)$ gives: $$\backslash begin\{align\}$$

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}
     

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Proof using the reciprocal rule or chain rule Let $h(x)\; =\; \backslash frac\{f(x)\}\{g(x)\}\; =\; f(x)\; \backslash cdot\; \backslash frac\{1\}\{g(x)\}.$

Then the product rule gives $h\text{'}(x)\; =\; f\text{'}(x)\backslash cdot\backslash frac\{1\}\{g(x)\}\; +\; f(x)\; \backslash cdot\; \backslash frac\{d\}\{dx\}\backslash left\backslash frac\{1\}\{g(x)\}\backslash right.$

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

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

  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}
     

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Proof by logarithmic differentiation Let $h(x)=\backslash frac\{f(x)\}\{g(x)\}.$ Taking the absolute value and natural logarithm of both sides of the equation gives $$\backslash ln|h(x)|=\backslash ln\backslash left|\backslash frac\{f(x)\}\{g(x)\}\backslash right|$$

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

Taking the logarithmic derivative of both sides, $$\backslash frac\{h\text{'}(x)\}\{h(x)\}=\backslash frac\{f\text{'}(x)\}\{f(x)\}-\backslash frac\{g\text{'}(x)\}\{g(x)\}$$

Solving for $h\text{'}(x)$ and substituting back $\backslash tfrac\{f(x)\}\{g(x)\}$ for $h(x)$ gives:

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

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Higher order derivatives Implicit differentiation can be used to compute the th derivative of a quotient (partially in terms of its first 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}.

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See also

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