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 $h(x)=\frac$, where both and are differentiable and $g(x)\neq 0.$ The quotient rule states that the derivative of is
:$h'(x) = \frac.$

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

Examples

Example 1: Basic example

Given $h(x)=\frac$, let $f(x)=e^x, g(x)=x^2$, then using the quotient rule:$\begin \frac \left(\frac\right) &= \frac \\ &= \frac \\ &= \frac \\ &= \frac \\ &= \frac. \end$

Example 2: Derivative of tangent function

The quotient rule can be used to find the derivative of $\tan x = \frac$ as follows:
$\begin \frac \tan x &= \frac \left(\frac\right) \\ &= \frac \\ &= \frac \\ &= \frac \\ &= \frac = \sec^2 x. \end$

Reciprocal rule

The reciprocal rule is a special case of the quotient rule in which the numerator $f(x)=1$. Applying the quotient rule givesh'(x)=\frac\left frac\right\frac=\frac.

Utilizing the chain rule yields the same result.

Proofs

Proof from derivative definition and limit properties

Let $h(x) = \frac.$ 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
h'(x) &= \lim_ \frac \\
&= \lim_ \frac \\
&= \lim_ \frac \\
&= \lim_ \frac \cdot \lim_\frac \\
&= \lim_ \left frac \right\cdot \frac \\
&= \left lim_ \frac - \lim_\frac \right\cdot \frac \\
&= \left lim_ \frac \cdot g(x) - f(x) \cdot \lim_\frac \right\cdot \frac \\
&= \frac.
\endThe limit evaluation $\lim_\frac=\frac$ is justified by the differentiability of $g(x)$, implying continuity, which can be expressed as $\lim_g(x+k) = g(x)$.

Proof using implicit differentiation

Let $h(x) = \frac,$ 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:
$\begin h'(x) &= \frac \\ &= \frac \\ &= \frac. \end$

Proof using the reciprocal rule or chain rule

Let $h(x) = \frac = f(x) \cdot \frac.$

Then the product rule gives h'(x) = f'(x)\cdot\frac + f(x) \cdot \frac\left frac\right

To evaluate the derivative in the second term, apply the reciprocal rule, or the power rule along with the chain rule:
\frac\left frac\right= -\frac \cdot g'(x) = \frac.

Substituting the result into the expression gives\begin
h'(x) &= f'(x)\cdot\frac + f(x)\cdot\left frac\right\\

&= \frac - \frac \\

&= \cdot - \frac \\

&= \frac.
\end

Proof by logarithmic differentiation

Let $h(x)=\frac.$ Taking the absolute value and natural logarithm of both sides of the equation gives
$\ln, h(x), =\ln\left, \frac\$

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

Taking the logarithmic derivative of both sides,
$\frac=\frac-\frac$

Solving for $h'(x)$ and substituting back $\tfrac$ for $h(x)$ gives:
\begin
h'(x)&=h(x)\left frac-\frac\right\
&=\frac\left frac-\frac\right\
&=\frac-\frac\\
&=\frac.
\end

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 $\tfrac(\ln, u, )=\tfrac$, which justifies taking the absolute value of the functions for logarithmic differentiation.

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 $f'' = g''h + 2g'h' + gh''$) and then solving for $h''$ yields$h'' = \left(\frac\right)'' = \frac.$

See also

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References

{{Calculus topics

Articles containing proofs
Differentiation rules
Theorems in mathematical analysis
Theorems in calculus