calculus Calculus, originally called infinitesimal calculus or "the calculus of infinitesimals", is the mathematical study of continuous change, in the same way that geometry is the study of shape, and algebra is the study of generalizations of arithm ...

, the quotient rule is a method of finding the

derivative In mathematics, the derivative of a function of a real variable measures the sensitivity to change of the function value (output value) with respect to a change in its argument (input value). Derivatives are a fundamental tool of calculus. F ...

of a

function Function or functionality may refer to: Computing * Function key, a type of key on computer keyboards * Function model, a structured representation of processes in a system * Function object or functor or functionoid, a concept of object-oriente ...

that is the ratio of two differentiable functions. Let

h(x)=f(x)/g(x),

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: Derivatives of tangent and cotangent functions

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

Similarly, the derivative of

\cot x = \frac

can be obtained as follows:

\begin
    \frac \cot x &= \frac \left(\frac\right) \\
    &= \frac \\
    &= \frac \\
    &= \frac \\
    &= \frac = -\csc^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 gives

\frac=\frac.

The result can also be derived using the definition of the derivative with limits.

\\ &= -g'(x) \cdot \frac \\ &= \frac. \end

The 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)

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:

\cdot \frac \\ &= \frac. \end

The 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,

f(x) = g(x)h(x).

The

product rule In calculus, the product rule (or Leibniz rule or Leibniz product rule) is a formula used to find the derivatives of products of two or more functions. For two functions, it may be stated in Lagrange's notation as (u \cdot v)' = u ' \cdot v + ...

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 In calculus, the reciprocal rule gives the derivative of the reciprocal of a function ''f'' in terms of the derivative of ''f''. The reciprocal rule can be used to show that the power rule holds for negative exponents if it has already been e ...

, or the

power rule In calculus, the power rule is used to differentiate functions of the form f(x) = x^r, whenever r is a real number. Since differentiation is a linear operation on the space of differentiable functions, polynomials can also be differentiated usin ...

along with the

chain rule In calculus, the chain rule is a formula that expresses the derivative of the composition of two differentiable functions and in terms of the derivatives of and . More precisely, if h=f\circ g is the function such that h(x)=f(g(x)) for every , ...

= -\frac \cdot g'(x) = \frac.

Substituting the result into the expression gives

\\ &= \frac - \frac \\ &= \frac. \end

Proof by logarithmic differentiation

Let

h(x)=\frac.

Taking the absolute value and

natural logarithm The natural logarithm of a number is its logarithm to the base of the mathematical constant , which is an irrational and transcendental number approximately equal to . The natural logarithm of is generally written as , , or sometimes, if ...

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 In mathematics, specifically in calculus and complex analysis, the logarithmic derivative of a function ''f'' is defined by the formula \frac where f' is the derivative of ''f''. Intuitively, this is the infinitesimal relative change in ''f''; ...

of both sides,

\frac=\frac-\frac

Solving for

h'(x)

and substituting back

f(x)/g(x)

for

h(x)

gives:

\ &=\frac-\frac\\ &=\frac. \end

Note: Taking the absolute value of the functions is necessary to allow

logarithmic differentiation In calculus, logarithmic differentiation or differentiation by taking logarithms is a method used to differentiate functions by employing the logarithmic derivative of a function ''f'', :(\ln f)' = \frac \quad \implies \quad f' = f \cdot (\ ...

of functions that can have negative values, as logarithms are only defined 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.

References

{{Calculus topics Articles containing proofs Differentiation rules Theorems in analysis Theorems in calculus