Quotient rule

Topics in Calculus

Fundamental theorem Limits of functions Continuity Mean value theorem Differential calculus  Derivative Change of variables Implicit differentiation Taylor's theorem Related rates Rules and identities: Power rule, Product rule, Quotient rule, Chain rule Integral calculus  Integral Lists of integrals Improper integrals Integration by: parts, disks, cylindrical shells, substitution, trigonometric substitution, partial fractions, changing order Vector calculus  Gradient Divergence Curl Laplacian Gradient theorem Green's theorem Stokes' theorem Divergence theorem Multivariable calculus  Matrix calculus Partial derivative Multiple integral Line integral Surface integral Volume integral Jacobian

In calculus, the quotient rule is a method of finding the derivative of a function that is the quotient of two other functions for which derivatives exist.

If the function one wishes to differentiate, f(x), can be written as

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

and $h(x)\not=0$, then the rule states that the derivative of g(x) / h(x) is

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

More precisely, if all x in some open set containing the number a satisfy $h(x)\not=0$, and g'(a) and h'(a) both exist, then f'(a) exists as well and

$f'(a)=\frac{h(a)g'(a) - h'(a)g(a)}{[h(a)]^2}.$

And this can be extended to calculate the second derivative as well (you can prove this by taking the derivative of f(x) = g(x)(h(x)) ^{− 1} twice). The result of this is:

$f''(x)=\frac{g''(x)[h(x)]^2-2g'(x)h(x)h'(x)+g(x)[2[h'(x)]^2-h(x)h''(x)]}{[h(x)]^3}.$

The quotient rule formula can be derived from the product rule and chain rule.

Examples

The derivative of (4x − 2) / (x² + 1) is:

$\begin{align}\frac{d}{dx}\left[\frac{(4x - 2)}{x^2 + 1}\right] &= \frac{(x^2 + 1)(4) - (4x - 2)(2x)}{(x^2 + 1)^2}\\ & = \frac{(4x^2 + 4) - (8x^2 - 4x)}{(x^2 + 1)^2} = \frac{-4x^2 + 4x + 4}{(x^2 + 1)^2}\end{align}$

In the example above, the choices

g(x) = 4x − 2

h(x) = x² + 1

were made. Analogously, the derivative of sin(x)/x² (when x ≠ 0) is:

$\frac{\cos(x) x^2 - \sin(x)2x}{x^4}$

Another example is:

$f(x) = \frac{2x^2}{x^3}$

whereas g(x) = 2x² and h(x) = x³, and g'(x) = 4x and h'(x) = 3x².

The derivative of f(x) is determined as follows:

$f'(x) = \frac {\left(4x \cdot x^3 \right) - \left(2x^2 \cdot 3x^2 \right)} {\left(x^3\right)^2} = \frac{4x^4 - 6x^4}{x^6} = \frac{-2x^4}{x^6} = -\frac{2}{x^2}$

This can be checked by using laws of exponents and the power rule:

$f(x) = \frac{2x^2}{x^3} = \frac{2}{x} = 2x^{-1}$

$f'(x) = -2x^{-2} = -\frac{2}{x^2}$

Limitations

The quotient rule is not useful at points where either the numerator or denominator are not differentiable; it's possible that the quotient may be differentiable at such points. For example, consider the function:

$f(x) = \frac{|x|+1}{|x|+1},$

where |x| denotes the absolute value of x. This is, of course, simply the function f(x) = 1, so it is differentiable everywhere and in particular f'(0) = 0. If we try to use the quotient rule to compute f'(0), however, an undefined value will result, since |x| is nondifferentiable at x = 0.

Proofs

Algebraic proof

Link to external website

From Newton's difference quotient

Suppose f(x) = g(x) / h(x) where $h(x) \neq 0$ and g and h are differentiable.

$f'(x) = \lim_{\Delta x \to 0} \frac{f(x + \Delta x)- f(x)}{\Delta x} = \lim_{\Delta x \to 0} \frac{\frac{g(x + \Delta x)}{h(x + \Delta x)} - \frac{g(x)}{h(x)}}{\Delta x}$

We pull out the 1 / Δx and combine the fractions in the numerator:

$= \lim_{\Delta x \to 0} \frac{1}{\Delta x} \left(\frac{g(x+\Delta x)h(x)-g(x)h(x+\Delta x)}{h(x)h(x+\Delta x)} \right)$

Adding and subtracting g(x)h(x) in the numerator:

$= \lim_{\Delta x \to 0} \frac{1}{\Delta x} \left( \frac{g(x+\Delta x)h(x)-g(x)h(x)-g(x)h(x+\Delta x)+g(x)h(x)}{h(x)h(x+\Delta x)} \right)$

We factor this and multiply the 1 / Δx through the numerator:

$= \lim_{\Delta x \to 0} \frac{\frac{g(x+\Delta x)-g(x)}{\Delta x}h(x)-g(x)\frac{h(x+\Delta x)-h(x)}{\Delta x}}{h(x)h(x+\Delta x)}$

Now we move the limit through:

$= \frac{\lim_{\Delta x \to 0} \left(\frac{g(x+\Delta x)-g(x)}{\Delta x}\right)h(x) - g(x) \lim_{\Delta x \to 0} \left(\frac{h(x+\Delta x)-h(x)}{\Delta x}\right)}{h(x)\lim_{\Delta x \to 0}h( x+\Delta x)}$

By the definition of the derivative, the limits of difference quotients in the numerator are derivatives. The limit in the denominator is h(x) because differentiable functions are continuous. Thus we get:

$= \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2}$

Using the chain rule

Consider the identity

$\frac{u}{v}\; =\; \frac{1}{4}\left[ \left( u+\frac{1}{v} \right)^{2}-\; \left( u-\frac{1}{v} \right)^{2} \right]$

Then

$\frac{d\left( \frac{u}{v} \right)}{dx}\; =\; \frac{d}{dx}\frac{1}{4}\left[ \left( u+\frac{1}{v} \right)^{2}-\; \left( u-\frac{1}{v} \right)^{2} \right]$

Leading to

$\frac{d\left( \frac{u}{v} \right)}{dx}\; =\; \frac{1}{4}\left[ 2\left( u+\frac{1}{v} \right)\left( \frac{du}{dx}-\frac{dv}{v^{2}dx} \right)-\; 2\left( u-\frac{1}{v} \right)\left( \frac{du}{dx}+\frac{dv}{v^{2}dx} \right) \right]$

Multiplying out leads to

$\frac{d\left( \frac{u}{v} \right)}{dx}\; =\; \frac{1}{4}\left[ \frac{4}{v}\frac{du}{dx}-\frac{4u}{v^{2}}\frac{dv}{dx} \right]$

Finally, taking a common denominator leaves us with the expected result

$\frac{d\left( \frac{u}{v} \right)}{dx}\; =\; \frac{\left[ v\frac{du}{dx}-u\frac{dv}{dx} \right]}{v^{2}}$

Using the product rule

Let $y = \frac{u}{v}$

Rewriting in indices form

$y = \frac{u}{v} = uv^{-1}.$

Using the product rule and the chain rule to differentiate leads us to,

$\frac{dy}{dx} = u'v^{-1} - v^{-2}uv' = \frac{u'}{v} - \frac{uv'}{v^{2}}$

Multiplying the first fraction's numerator and denominator with v we get,

$\frac{dy}{dx} = \frac{vu'}{v^2} - \frac{uv'}{v^{2}} = \frac{vu' - uv'}{v^{2}}$

Which is the quotient rule.

By total differentials

An even more elegant proof is a consequence of the law about total differentials, which states that the total differential,

$dF = \frac{\partial F}{\partial x} dx + \frac{\partial F}{\partial y} dy + \frac{\partial F}{\partial z} dz + \cdots$

of any function in any set of quantities is decomposable in this way, no matter what the independent variables in a function are (i.e., no matter which variables are taken so that they may not be expressed as functions of other variables). This means that, if N and D are both functions of an independent variable x, and F = N(x)/D(x), then it must be true both that

$(*) \qquad dF = \frac{\partial F}{\partial x} \, dx$

and that

$dF = \frac{\partial F}{\partial N}dN + \frac{\partial F}{\partial D} \, dD.$

But we know that dN = N'(x)dx and $dD = D'(x) \, dx.$

Substituting and setting these two total differentials equal to one another (since they represent limits which we can manipulate), we obtain the equation

$\frac{\partial F}{\partial x} dx = \frac{\partial F}{\partial N}N'(x) dx + \frac{\partial F}{\partial D}D'(x) dx$

which requires that

$(\#) \qquad \frac{\partial F}{\partial x} = \frac{\partial F}{\partial N}N'(x) + \frac{\partial F}{\partial D}D'(x).$

We compute the partials on the right:

$\frac{\partial F}{\partial N} = \frac{\partial (N/D)}{\partial N} = \frac{1}{D};$

$\frac{\partial F}{\partial D} = \frac{\partial (N/D)}{\partial D} = -\frac{N}{D^2}.$

If we substitute them into (#),

$\frac{\partial F}{\partial x} = \frac{N'(x)}{D(x)} - \frac{N(x) D'(x)}{D(x)^2}$

$\frac{\partial F}{\partial x} = \frac{D(x)N'(x)}{D(x)^2} - \frac{N(x) D'(x)}{D(x)^2}$

which gives us the quotient rule, since, by (*),

$\frac{dF}{dx} = \frac{\partial F}{\partial x}.$

This proof, of course, is just another, more systematic (even if outmoded) way of proving the theorem in terms of limits, and is therefore equivalent to the first proof above – and even reduces to it, if you make the right substitutions in the right places. Students of multivariable calculus will recognize it as one of the chain rules for functions of multiple variables.

See also

• Chain rule
• Difference quotient
• Product rule
• Reciprocal rule