1. ## Even/odd function differentiability

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

2. Originally Posted by frenchguy87
How would you prove that if $f$ is an even function ( $f(-x)=f(x)$ ) and has a derivative at every point, then the derivative $f'$ is an odd function ( $f'(-x) = -f'(x)$ ) ?

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

Tonio

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

Tonio
How do you get to $f'(-x)=-f'(x)$ from that ?

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

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

Tonio

5. I don't know why but I'm not convinced

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

Perhaps because you don't understand...?

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

Tonio

7. 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.