Editing Quotient rule

{{short description|Formula for the derivative of a ratio of functions}}
{{Calculus |Differential}}
In [[calculus]], the '''quotient rule''' is a method of finding the [[derivative]] of a [[function (mathematics)|function]] that is the ratio of two differentiable functions. Let <math>h(x)=\frac{f(x)}{g(x)}</math>, where both {{mvar|f}} and {{mvar|g}} are differentiable and <math>g(x)\neq 0.</math> The quotient rule states that the derivative of {{math|''h''(''x'')}} 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 [[Differentiation rules|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 ==
{{Main|Reciprocal rule}}
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 function|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 {{mvar|n}}th derivative of a quotient (partially in terms of its first {{math|''n'' &minus; 1}} derivatives).  For example, differentiating <math>f=gh</math> twice (resulting in <math>f'' = g''h + 2g'h' + gh''</math>) and then solving for <math>h''</math> yields<math display="block">h'' = \left(\frac{f}{g}\right)'' = \frac{f''-g''h-2g'h'}{g}.</math>
==See also==

* {{annotated link|Chain rule}}
* {{annotated link|Differentiation of integrals}}
* {{annotated link|Differentiation rules}}
* {{annotated link|General Leibniz rule}}
* {{annotated link|Inverse functions and differentiation}}
* {{annotated link|Linearity of differentiation}}
* {{annotated link|Product rule}}
* {{annotated link|Reciprocal rule}}
* {{annotated link|Table of derivatives}}
* {{annotated link|Vector calculus identities}}

==References==

{{reflist}}

{{Calculus topics}}

[[Category:Articles containing proofs]]
[[Category:Differentiation rules]]
[[Category:Theorems in mathematical analysis]]
[[Category:Theorems in calculus]]