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In calculus, the quotient rule is a method of finding the derivative of a function that is the ratio of two differentiable functions.^[1][2][3] Let ${\displaystyle h(x)={\frac {f(x)}{g(x)))}$, where both f and g are differentiable and ${\displaystyle g(x)\neq 0.}$ The quotient rule states that the derivative of h(x) is

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

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

Examples

Example 1: Basic example

Given ${\displaystyle h(x)={\frac {e^{x)){x^{2))))$, let ${\displaystyle f(x)=e^{x},g(x)=x^{2))$, then using the quotient rule:

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

Example 2: Derivative of tangent function

The quotient rule can be used to find the derivative of ${\displaystyle \tan x={\frac {\sin x}{\cos x))}$ as follows:

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

Reciprocal rule

The reciprocal rule is a special case of the quotient rule in which the numerator ${\displaystyle f(x)=1}$. Applying the quotient rule gives

$${\displaystyle h'(x)={\frac {d}{dx))\left[{\frac {1}{g(x)))\right]={\frac {0\cdot g(x)-1\cdot g'(x)}{g(x)^{2))}={\frac {-g'(x)}{g(x)^{2))}.}$$

Utilizing the chain rule yields the same result.

Proofs

Proof from derivative definition and limit properties

Let ${\displaystyle h(x)={\frac {f(x)}{g(x))).}$ Applying the definition of the derivative and properties of limits gives the following proof, with the term ${\displaystyle f(x)g(x)}$ added and subtracted to allow splitting and factoring in subsequent steps without affecting the value:

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

${\displaystyle \lim _{k\to 0}{\frac {1}{g(x+k)g(x)))={\frac {1}{g(x)^{2))))$

${\displaystyle g(x)}$

${\displaystyle \lim _{k\to 0}g(x+k)=g(x)}$

Proof using implicit differentiation

Let ${\displaystyle h(x)={\frac {f(x)}{g(x))),}$ so that ${\displaystyle f(x)=g(x)h(x).}$

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

Solving for ${\displaystyle h'(x)}$ and substituting back for ${\displaystyle h(x)}$ gives:

$${\displaystyle {\begin{aligned}h'(x)&={\frac {f'(x)-g'(x)h(x)}{g(x)))\\&={\frac {f'(x)-g'(x)\cdot {\frac {f(x)}{g(x)))}{g(x)))\\&={\frac {f'(x)g(x)-f(x)g'(x)}{g(x)^{2))}.\end{aligned))}$$

Proof using the reciprocal rule or chain rule

Let ${\displaystyle h(x)={\frac {f(x)}{g(x)))=f(x)\cdot {\frac {1}{g(x))).}$

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

To evaluate the derivative in the second term, apply the reciprocal rule, or the power rule along with the chain rule:

$${\displaystyle {\frac {d}{dx))\left[{\frac {1}{g(x)))\right]=-{\frac {1}{g(x)^{2))}\cdot g'(x)={\frac {-g'(x)}{g(x)^{2))}.}$$

Substituting the result into the expression gives

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

Proof by logarithmic differentiation

Let ${\displaystyle h(x)={\frac {f(x)}{g(x))).}$ Taking the absolute value and natural logarithm of both sides of the equation gives

$${\displaystyle \ln |h(x)|=\ln \left|{\frac {f(x)}{g(x)))\right|}$$

Applying properties of the absolute value and logarithms,

$${\displaystyle \ln |h(x)|=\ln |f(x)|-\ln |g(x)|}$$

Taking the logarithmic derivative of both sides,

$${\displaystyle {\frac {h'(x)}{h(x)))={\frac {f'(x)}{f(x)))-{\frac {g'(x)}{g(x)))}$$

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

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

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 ${\displaystyle {\tfrac {d}{dx))(\ln |u|)={\tfrac {u'}{u))}$, 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 ${\displaystyle f=gh}$ twice (resulting in ${\displaystyle f''=g''h+2g'h'+gh''}$) and then solving for ${\displaystyle h''}$ yields

$${\displaystyle h''=\left({\frac {f}{g))\right)''={\frac {f''-g''h-2g'h'}{g)).}$$