Prove that the derivative of an odd function is even

[NWeid1]

Homework Statement

A function f is an even function is f(-x)=f(x) for all x and is an odd function is f(-x)=-f(x) for all x. Prove that the derivative of an odd function is even and the derivative of an even function is off. I get what even and odd functions are but I'm not sure how to rigorously prove this.

Homework Equations

The Attempt at a Solution

Not sure how to start. Maybe use limit definition?

 

[hunt_mat]

Write the definition of the derivative as:
$$f'(a)=\lim_{x\rightarrow a}\frac{f(x)-f(a)}{x-a}$$
and apply the definition of an odd function.

 

[Dick]

Homework Statement

A function f is an even function is f(-x)=f(x) for all x and is an odd function is f(-x)=-f(x) for all x. Prove that the derivative of an odd function is even and the derivative of an even function is off. I get what even and odd functions are but I'm not sure how to rigorously prove this.

Homework Equations

The Attempt at a Solution

Not sure how to start. Maybe use limit definition?

Yes, use the limit definition of derivative. Now try and get started.

 

[Dick]

Write the definition of the derivative as:
$$f'(x)=\lim_{x\rightarrow a}\frac{f(x)-f(a)}{x-a}$$
and apply the definition of an odd function.

That's f'(a), right?

 

[NWeid1]

I just don't understand how to use limit definition of just f(x) and not an actualy given function.

 

[hunt_mat]

That's f'(a), right?

Yes, changed.

 

[Curious3141]

Homework Statement

A function f is an even function is f(-x)=f(x) for all x and is an odd function is f(-x)=-f(x) for all x. Prove that the derivative of an odd function is even and the derivative of an even function is off. I get what even and odd functions are but I'm not sure how to rigorously prove this.

Homework Equations

The Attempt at a Solution

Not sure how to start. Maybe use limit definition?

Chain rule is the easiest way, if you're allowed to use it. What's the derivative of (-x)? Work with that.

 

[NWeid1]

Yeah, I thought about using chain rule, but how could I prove it using the chain rule? I am just so confused by this concept -.- lol

 

[Curious3141]

Yeah, I thought about using chain rule, but how could I prove it using the chain rule? I am just so confused by this concept -.- lol

The chain rule is essentially about differentiating the composition of two (or more) functions - meaning functions applied "one on top of another".

Say you have the composite function f(g(x)). Let's put g(x) = y

By Chain Rule,

f'(g(x)) = df/dx = (df/dy)*(dy/dx) = f'(y)*g'(x) = f'(g(x))*g'(x)

Now put g(x) = -x. What happens to f'(g(x))?

For an odd function, f(-x) = -f(x). What's the relationship between f'(-x) and f'(x)?

For an even function, f(-x) = f(x). What's the relationship between f'(-x) and f'(x)?

 

[Dick]

Yeah, I thought about using chain rule, but how could I prove it using the chain rule? I am just so confused by this concept -.- lol

If you can get the even or odd case using the chain rule, then you can probably figure out how to do it from the limit definition. It's not that different. You use the same idea.

 

[NWeid1]

I think I got it, and it was so much easier than what I thought.

For an even function:
f(x) = f(-x)
Take the derivative of each side (chain rule for f(-x))
f'(x) = -f(-x)
f(-x) = -f(x), therefore it is an off function.

and so the same process for an odd function. Am I right?

 

[Dick]

I think I got it, and it was so much easier than what I thought.

For an even function:
f(x) = f(-x)
Take the derivative of each side (chain rule for f(-x))
f'(x) = -f(-x)
f(-x) = -f(x), therefore it is an off function.

and so the same process for an odd function. Am I right?

You probably would be if you weren't being so sloppy. You meant f'(x)=-f'(-x). I hope.

 

[NWeid1]

Lol. Yeah. My B