# Prove that f'(0) = 0

1. Jun 16, 2011

### Glissando

1. The problem statement, all variables and given/known data
If f(x) is an even function and f'(x) exists for all x, prove that f'(0) = 0. (Hint: Start with an equation that is true for all even functions and differentiate both sides with respect to x.)

2. Relevant equations
Equation true for all even functions: f(x) = f(-x)

3. The attempt at a solution

f(x) = f(-x)

f(d/dx (x)) = f(d/dx (-x))

f(1) = f(-1)

I'm not sure if I have the notation correct when differentiating...or I could have done something like this:

f(x) = f(-x)

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

f'(0) = f'(-0)

Then I'm not too sure as to what to do from there ):

Thanks for all the help <3

2. Jun 16, 2011

### Dick

The derivative of f(x) is f'(x), sure. The derivative of f(-x) isn't equal to f'(-x). You need to use the chain rule to differentiate f(-x).

3. Jun 16, 2011

### Glissando

But I'm taking the derivative with respect to x.... ): why do I need to use the chain rule?

4. Jun 16, 2011

### Dick

Because f(-x) has the form f(g(x)) where g(x)=(-x). Doesn't it?

5. Jun 16, 2011

### Glissando

IF I am doing this right....

f(x) = f(-x)

f' (x) = f'(-x) + f(-1) derivative of outer function times inner function plus outside function times derivative of inside function ?

Thank you for your patience (:!

6. Jun 16, 2011

### eumyang

No, I think you're confusing with the product rule.

7. Jun 16, 2011

### Dick

d/dx of f(g(x)) is f'(g(x))*g'(x), right? Isn't that the chain rule? If you put g(x)=(-x) what do you get for the derivative of f(-x)??

8. Jun 16, 2011

### Glissando

Goodness I hope I'm doing it right this time, thanks for bearing with me.

f'(x) = f'(g(x)) * g'(x)

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

f'(x) = -f'(-x).....= f'(x)?

God I'm feeling so dumb right now ):!!!!

9. Jun 16, 2011

### Dick

You shouldn't feel dumb now that you are getting it right. Sure, f'(x)=(-f'(-x)). Put x=0 and tell me what f'(0) must be.

10. Jun 16, 2011

### Glissando

God you're good <3

Thank you so much (:!!!!

11. Jun 17, 2011

### HallsofIvy

Staff Emeritus
Of course, he is!