Quotient rule

Part of a series of articles about
Calculus
${\displaystyle \int _{a}^{b}f'(t)\,dt=f(b)-f(a)}$
Fundamental theorem Limits Continuity Rolle's theorem Mean value theorem Inverse function theorem
Differential Definitions Derivative (generalizations) Differential infinitesimal of a function total Concepts Differentiation notation Second derivative Implicit differentiation Logarithmic differentiation Related rates Taylor's theorem Rules and identities Sum Product Chain Power Quotient L'Hôpital's rule Inverse General Leibniz Faà di Bruno's formula Reynolds
Integral Lists of integrals Integral transform Leibniz integral rule Definitions Antiderivative Integral (improper) Riemann integral Lebesgue integration Contour integration Integral of inverse functions Integration by Parts Discs Cylindrical shells Substitution (trigonometric, tangent half-angle, Euler) Euler's formula Partial fractions Changing order Reduction formulae Differentiating under the integral sign Risch algorithm
Series Geometric (arithmetico-geometric) Harmonic Alternating Power Binomial Taylor Convergence tests Summand limit (term test) Ratio Root Integral Direct comparison Limit comparison Alternating series Cauchy condensation Dirichlet Abel
Vector Gradient Divergence Curl Laplacian Directional derivative Identities Theorems Gradient Green's Stokes' Divergence generalized Stokes Helmholtz decomposition
Multivariable Formalisms Matrix Tensor Exterior Geometric Definitions Partial derivative Multiple integral Line integral Surface integral Volume integral Jacobian Hessian
Advanced Calculus on Euclidean space Generalized functions Limit of distributions
Specialized Fractional Malliavin Stochastic Variations
Miscellaneous Precalculus History Glossary List of topics Integration Bee Mathematical analysis Nonstandard analysis
v t e

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:

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:

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

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:

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

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:

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:

Substituting the result into the expression gives

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

Applying properties of the absolute value and logarithms,

Taking the logarithmic derivative of both sides,

Solving for ${\displaystyle h'(x)}$ and substituting back ${\displaystyle {\tfrac {f(x)}{g(x)}}}$ for ${\displaystyle h(x)}$ 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 ${\displaystyle {\tfrac {d}{dx}}(\ln |u|)={\tfrac {u'}{u}}}$, 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

See also

• Chain rule – For derivatives of composed functions
• Differentiation of integrals – Problem in mathematics
• Differentiation rules – Rules for computing derivatives of functions
• General Leibniz rule – Generalization of the product rule in calculus
• Inverse functions and differentiation – Calculus identityPages displaying short descriptions of redirect targets
• Linearity of differentiation – Calculus property
• Product rule – Formula for the derivative of a product
• Reciprocal rule – differentiation rulePages displaying wikidata descriptions as a fallback
• Table of derivatives – Rules for computing derivatives of functionsPages displaying short descriptions of redirect targets
• Vector calculus identities – Mathematical identities

References