# Quotient rule

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.[1][2][3]

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

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

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 $(4x - 2)/(x^2 + 1)$ is:

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

$g(x) = 4x - 2$
$h(x) = x^2 + 1$

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

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

## Proof

Let some function $f(x)=\frac{g(x)}{h(x)}$.
We wish to find $\frac{d}{dx}[f(x)]$, and this is equivalent to $\frac{d}{dx}\left[\frac{g(x)}{h(x)}\right]$.
We also know that $g(x)=f(x)\cdot h(x)$.
By the product rule, we can say that $\frac{d}{dx}[g(x)]=h(x)\cdot\frac{d}{dx}[f(x)]+\frac{d}{dx}[h(x)]\cdot f(x)$.
From the equation above, $\frac{d}{dx}[f(x)]=\frac{\frac{d}{dx}[g(x)]-\frac{d}{dx}[h(x)]\cdot f(x)}{h(x)}$
Because $f(x)=\frac{g(x)}{h(x)}$, the right side simplifies to $\frac{\frac{d}{dx}[g(x)]-\frac{d}{dx}[h(x)]\cdot\frac{g(x)}{h(x)}}{h(x)}$, which simplifies to $\frac{d}{dx}[f(x)]=\frac{\frac{d}{dx}[g(x)]\cdot h(x)-\frac{d}{dx}[h(x)]\cdot g(x)}{[h(x)]^2}$
Hence proved. [4]

## References

1. ^ Stewart, James (2008). Calculus: Early Transcendentals (6th ed.). Brooks/Cole. ISBN 0-495-01166-5.
2. ^ Larson, Ron; Edwards, Bruce H. (2009). Calculus (9th ed.). Brooks/Cole. ISBN 0-547-16702-4.
3. ^ Thomas, George B.; Weir, Maurice D.; Hass, Joel (2010). Thomas' Calculus: Early Transcendentals (12th ed.). Addison-Wesley. ISBN 0-321-58876-2.
4. ^ Berresford, Geoffrey; Rockett, Andrew (2008). Brief Applied Calculus (5th ed.). ISBN 0-547-16977-9.