1. ## even and odd

Recall that $f: \mathbb{R} \to \mathbb{R}$ is an even function if $f(-x)=f(x) , \ \forall x \in \mathbb{R}$
and is an odd function if $f(-x)=-f(x) , \ \forall x \in \mathbb{R}$
prove that if $f: \mathbb{R} \to \mathbb{R}$ is an even differentiable function on $\mathbb{R} \$
then $f'$ is an odd function

2. Originally Posted by flower3
Recall that $f: \mathbb{R} \to \mathbb{R}$ is an even function if $f(-x)=f(x) , \ \forall x \in \mathbb{R}$
and is an odd function if $f(-x)=-f(x) , \ \forall x \in \mathbb{R}$
prove that if $f: \mathbb{R} \to \mathbb{R}$ is an even differentiable function on $\mathbb{R} \$
then $f'$ is an odd function

$f'(x_0):=\lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}$ ...and the quotient of even by odd functions is an odd function.

Tonio

3. Use the chain rule: the derivative of f(-x) is f'(-x)(-x)'= -f'(-x). The derivative of -f(x) is -f'(x). Put those together.

4. Originally Posted by tonio
$f'(x_0):=\lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}$ ...and the quotient of even by odd functions is an odd function.

Tonio
I don't understand your point, tonio. If $g(x)= x- x_0$, then $g(-x)= -x-x_0$ is not equal to either $g(x)= x-x_0$ nor $-g(x)= -x+ x_0$ so g is neither even nor odd. What even and odd functions make up your quotient?

5. Originally Posted by HallsofIvy
I don't understand your point, tonio. If $g(x)= x- x_0$, then $g(-x)= -x-x_0$ is not equal to either $g(x)= x-x_0$ nor $-g(x)= -x+ x_0$ so g is neither even nor odd. What even and odd functions make up your quotient?

I wrote that too quick: I meant to write the definition of the derivative as

$f'(x_0)=\lim_{h\to 0}\frac{f(x_0+h)-f(x_0)}{h}$ , and then

$f'(-x_0)=\lim_{h\to 0}\frac{f(-x_0+h)-f(-x_0)}{h}$ . Now, here substitute k:=-h, and of course $h\rightarrow 0 \Longleftrightarrow k\rightarrow 0$ , so then (and here enters the numerator even and the denominator odd thing)

$f'(-x_0)=\lim_{h\to 0}\frac{f(-x_0+h)-f(-x_0)}{h}=\lim_{k\to 0}\frac{f(-x_0-k)-f(-x_0)}{-k}=$ $\lim_{k\to 0}\frac{f(x_0+k)-f(x_0)}{-k}=-\lim_{k\to 0}\frac{f(x_0+k)-f(x_0)}{k}=-f'(x_0)$

Tonio

6. Originally Posted by flower3
Recall that $f: \mathbb{R} \to \mathbb{R}$ is an even function if $f(-x)=f(x) , \ \forall x \in \mathbb{R}$
and is an odd function if $f(-x)=-f(x) , \ \forall x \in \mathbb{R}$
prove that if $f: \mathbb{R} \to \mathbb{R}$ is an even differentiable function on $\mathbb{R} \$
then $f'$ is an odd function
Essentially the same thing as HallsOfIvy is to define $h(x)=f(x)-f(-x)=0$ and then note that $h'(x)=f'(x)+f'(-x)=0\implies f'(x)=-f'(-x)$