Prove f'(0) = 0: An Even Function Homework Solution

[Glissando]

Homework Statement

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.)

Homework Equations

Equation true for all even functions: f(x) = f(-x)

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

 

[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).

 

[Glissando]

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).

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

 

[Dick]

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

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

 

[Glissando]

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

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 (:!

 

[eumyang]

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 (:!

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

 

[Dick]

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 (:!

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)??

 

[Glissando]

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)??

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 ):!

 

[Dick]

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 ):!

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.

 

[Glissando]

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.

God you're good <3

Thank you so much (:!

 

[HallsofIvy]

Of course, he is!