Reciprocal rule

From Wikipedia, the free encyclopedia

This is about a method in calculus. For other uses of "reciprocal", see reciprocal.

In calculus, the reciprocal rule is a shorthand method of finding the derivative of a function that is the reciprocal of a differentiable function, without using the quotient rule or chain rule.

The reciprocal rule states that the derivative of $1 / g (x)$ is given by

$\frac{d}{dx}\left(\frac{1}{g(x)}\right) = \frac{- g'(x)}{(g(x))^2}$

where $g(x) \neq 0.$

Contents

1 Proof
- 1.1 From the quotient rule
- 1.2 From the chain rule
2 Examples
3 See also

The reciprocal rule is derived from the quotient rule, with the numerator $f (x) = 1$ . Then,

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

The reciprocal rule can also be derived from the chain rule. Let $f (x) = x - 1 .$ Then,

$f(g(x))=(g(x))^{-1}=\frac{1}{g(x)}$ .

By the chain rule,

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

The derivative of $1 / (x 2 + 2 x)$ is:

$\frac{d}{dx}\left(\frac{1}{x^2 + 2x}\right) = \frac{-2x - 2}{(x^2 + 2x)^2}.$

The derivative of $1 / cos(x)$ (when $\cos x\not=0$ ) is:

$\frac{d}{dx} \left(\frac{1}{\cos(x) }\right) = \frac{-\sin(x)}{\cos^2(x)} = \frac{1}{\cos(x)} \frac{\sin(x)}{\cos(x)} = \sec(x)\tan(x).$

For more general examples, see the derivative article.

Retrieved from "http://en.wikipedia.org../../../r/e/c/Reciprocal_rule.html"

Category: Differential calculus

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