1. ## Using chain rule..

Use the chain rule to show that the derivative of an odd function is an even function.

Basically assuming that f(x) is odd:

f(-x) = -f(x) and f(x) = -f(x)

Show that:

f'(x) = f'(-x)

All must be done in terms of f. How would I go about in doing this?

$f(-x)=-f(x)$. Incidentally, the equation $f(x)=-f(x)$, if true for all $x$, implies that $f(x)=0$ for all $x$. Not sure that's where you want to go.

Try differentiating both sides of the initial equation. What does that give you?

3. Originally Posted by GameTheory

Use the chain rule to show that the derivative of an odd function is an even function.

Basically assuming that f(x) is odd:

f(-x) = -f(x) and f(x) = -f(x)

Show that:

f'(x) = f'(-x)

All must be done in terms of f. How would I go about in doing this?
Then chain rule says:

$\frac{d}{dx}[f(g(x))] = (f'(g(x)))(g'(x))$

So we have:

$f(x)$

and

$g(x) = -x$

Thus:

$f(g(x))= f(-x)$

so differentiating with the cain rule gets:

$\frac{d}{dx}[f(g(x))] = f'(-x) = [f'(g(x)][g'(x)]$

$= [f'(-x)]\frac{d}{dx}[-x] = -f'(-x)$

And since:

$-f(x) = f(x)$

we can move over the negitive in the original equation you gave to get:

$-f'(x) = f'(-x)$

and

$-f(x) = f(-x)$

thus:

$f'(x) = f'(-x)$

I feel like there is an error in this proof, but it seems logical, atleast I think I'm going in the right direction. Good luck.

4. f(-x) = -f(x)

f'(-x)*(-x)'=-f'(x)

f'(-x)*(-1)=-f'(x)

f'(-x)=f'(x)

Was that hard?