# Quotient rule

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vte :
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.

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

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

## 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.

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

### 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${\begin{aligned}h'(x)&=f'(x)\cdot {\frac {1}{g(x)}}+f(x)\cdot \left[{\frac {-g'(x)}{g(x)^{2}}}\right]\\&={\frac {f'(x)}{g(x)}}-{\frac {f(x)g'(x)}{g(x)^{2}}}\\&={\frac {g(x)}{g(x)}}\cdot {\frac {f'(x)}{g(x)}}-{\frac {f(x)g'(x)}{g(x)^{2}}}\\&={\frac {f'(x)g(x)-f(x)g'(x)}{g(x)^{2}}}.\end{aligned}}$

### 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: ${\begin{aligned}h'(x)&=h(x)\left[{\frac {f'(x)}{f(x)}}-{\frac {g'(x)}{g(x)}}\right]\\&={\frac {f(x)}{g(x)}}\left[{\frac {f'(x)}{f(x)}}-{\frac {g'(x)}{g(x)}}\right]\\&={\frac {f'(x)}{g(x)}}-{\frac {f(x)g'(x)}{g(x)^{2}}}\\&={\frac {f'(x)g(x)-f(x)g'(x)}{g(x)^{2}}}.\end{aligned}}$
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 nth derivative of a quotient (partially in terms of its first n − 1 derivatives). For example, differentiating $f=gh$ twice (resulting in $f''=g''h+2g'h'+gh''$) and then solving for $h''$ yields$h''=\left({\frac {f}{g}}\right)''={\frac {f''-g''h-2g'h'}{g}}.$