Even and Odd functions

1. Oct 22, 2004

JasonRox

If a function is even, prove that the derivative is odd.

Look at a graph of x^2 we can clearly see why.

This is how I would approach this...

If we solve d/dx x^n and n is an even integer, we get the derivative nx^(n-1). Since n is even, n-1 is odd.

Because n-1 is odd, the derivative nx^(n-1) becomes odd because f'(-x)= - f'(x). Therefore the derivative of an even function becomes an odd function.

Note: The TA couldn't solve this... :uhh:

I excluded the proof of d/dx x^n = nx^(n-1) because it is not necessary because I know how to do that.

2. Oct 22, 2004

NateTG

Since $$f(x)$$ is even we have:
$$f(x)=f(-x)$$
now, let's take the derivative of both sides. The LHS is easy:
$$\frac{d}{dx}f(x)=f'(x)$$
and we can do the RHS using the chain rule:
$$\frac{d}{dx}f(-x)=- f'(-x)$$
so, the original equation turns into:
$$f'(x)=-f'(-x)$$
multiply both sides by $$-1$$
$$-f'(x)=f'(-x)$$
which indicates that the function is odd.

3. Oct 22, 2004

HallsofIvy

Not all functions are powers of x!

If f(x)= sin(x), an odd function, its derivative is cos(x), an even function.

NateTG's response is the way to go.

4. Oct 22, 2004

JasonRox

Thanks NateTG.

Last edited: Oct 22, 2004
5. Feb 25, 2011

izilude

Yes but this would be sufficient if the function can be expanded in a power series. sin x = x - x^3/3! + x^5/5! +....

d sinx/dx = 1 - x^2/2! + x^4/4 +... = cos x.

Thus, his proof is correct for all analytic functions.