# Even/odd function differentiability

• Mar 22nd 2010, 03:40 AM
frenchguy87
Even/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 AM
tonio
Quote:

Originally Posted by frenchguy87
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)\$ ) ?

Apply the chain rule to \$\displaystyle f(-x):\,\,\,f'(-x)=-f'(-x)\$ , and thus ...

Tonio
• Mar 22nd 2010, 05:44 PM
frenchguy87
Quote:

Originally Posted by tonio
Apply the chain rule to \$\displaystyle f(-x):\,\,\,f'(-x)=-f'(-x)\$ , and thus ...

Tonio

How do you get to \$\displaystyle f'(-x)=-f'(x)\$ from that ?
• Mar 22nd 2010, 07:48 PM
tonio
Quote:

Originally Posted by frenchguy87
How do you get to \$\displaystyle f'(-x)=-f'(x)\$ from that ?

Well, since \$\displaystyle f(-x)=f(x)\Longrightarrow f'(-x)=f'(x)\$ , right?

Tonio
• Mar 24th 2010, 06:03 PM
frenchguy87
I don't know why but I'm not convinced
• Mar 24th 2010, 07:30 PM
tonio
Quote:

Originally Posted by frenchguy87
I don't know why but I'm not convinced

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 PM
Tinyboss
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.