Back to the index. Or to the chambers

This article has 2 links. View as Cloud or List.

1	Relation Theory
4	Chain Rule

Loading ...

Planetmath Browser (2008—2009)

BSD licence | A django site

All articles taken from PlanetMath.org snapshot under CC-BY-SA licence.

→ The original article on PlanetMath.org

Other Formats: LaTeX

New! You can click on formulas to copy the LaTeX source to your clipboard. | Math Videos

Derivative Of Evenodd Function Proof

Suppose $f(x)=\pm f(-x)$ . We need to show that $f'(x)=\mp f'(-x)$ . To do this, let us define the auxiliary function $m:\mathbb{R}\to \mathbb{R}$ , . The condition on is then $f(x) = \pm (f\circ m)(x)$ . Using the chain rule, we have that

$\displaystyle f'(x)$	$\displaystyle =$	$\displaystyle \pm(f\circ m)'(x)$
	$\displaystyle =$	$\displaystyle \pm f'\big(m(x)\big) m'(x)$
	$\displaystyle =$	$\displaystyle \mp f'(-x),$

and the claim follows. $\Box$