# A question on notation of derivatives

1. Jun 16, 2013

### Lemniscates

I was doing a proof on why the derivative of an even function is odd and vice versa. Now, the way I did the problem was by using the chain rule to rewrite the derivative of f(-x), and the proof worked out perfectly fine.

But I had a thought that I can't quite wrap my around, and I think it's just because I don't fully understand the notation. I almost made this mistake but caught myself because I realized it doesn't work; it's essentially a silly "proof" for why the derivative of an odd function is odd:

1. f(-x)=-f(x), the definition of an odd function
2. Taking the derivative of both sides (since two functions which are equal for all values of x should also have equivalent derivatives):
f'(-x)=-f'(x)

I know this is wrong (since that says the derivative is also odd). I'm also pretty certain that the derivative of -f(x) is -f'(x) (because of the constant rule). So that means that the derivative of f(-x) is NOT f'(-x). My question: why is the derivative of f(-x) not f'(-x)?

2. Jun 16, 2013

### jostpuur

In general a derivative of $f(g(x))$ is not $f'(g(x))$, so in particular there is no reason to assume, that a derivative of $f(-x)$ would be $f'(-x)$.

3. Jun 16, 2013

### mathman

You can use the chain rule. To illustrate, let u = -x, so f(-x) = f(u). df(u)/dx = f'(u)du/dx = -f'(-x).

4. Jun 16, 2013

### Lemniscates

Ah, I think I see now. So f'(-x) would be df/d(-x) (change of f over change of negative x as a single quantity), while (f(-x))' would be df/dx?