How would you prove that if $\displaystyle f$ is an even function ( $\displaystyle f(-x)=f(x)$ ) and has a derivative at every point, then the derivative $\displaystyle f'$ is an odd function ( $\displaystyle f'(-x) = -f'(x)$ ) ?

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- Mar 22nd 2010, 03:40 AMfrenchguy87Even/odd function differentiability
How would you prove that if $\displaystyle f$ is an even function ( $\displaystyle f(-x)=f(x)$ ) and has a derivative at every point, then the derivative $\displaystyle f'$ is an odd function ( $\displaystyle f'(-x) = -f'(x)$ ) ?

- Mar 22nd 2010, 04:16 AMtonio
- Mar 22nd 2010, 05:44 PMfrenchguy87
- Mar 22nd 2010, 07:48 PMtonio
- Mar 24th 2010, 06:03 PMfrenchguy87
I don't know why but I'm not convinced

- Mar 24th 2010, 07:30 PMtonio

Perhaps because you don't understand...?

If $\displaystyle f(-x)=f(x)$ then $\displaystyle (f(-x))'=f'(x)$ ; OTOH, applying the chain rule we have that $\displaystyle (f(-x))'=-f'(-x)$ and thus we get $\displaystyle f'(x)=-f'(-x)\Longleftrightarrow f'(-x)=-f'(x)$ ...

Tonio - Mar 24th 2010, 08:21 PMTinyboss
Tonio gave the true answer. If you need some intuition, just draw an even function (symmetric around the y-axis) and pick some point x. Then draw tangent lines at x and at -x. Clearly they have opposite slopes.