Differentiating f(-x)?

1. Apr 7, 2013

I came across this simple expression while doing some maths.

If $$\frac{d}{dx}f(x)=g(x)$$

Then $$\frac{d}{dx}f(-x)=-g(-x)$$

Is this correct? How do we prove it?

2. Apr 7, 2013

phyzguy

Do you know the chain rule? This is a simple example of it.

3. Apr 7, 2013

What are the specifics?

4. Apr 7, 2013

mathsciguy

Chain rule: $\frac{d[f(g(x))]}{dx} = \frac{d[f(g(x))]}{d[g(x)]} \frac{d[g(x)]}{dx}$. In your case, g(x) = -x.

5. Apr 7, 2013

jbunniii

If you prefer a direct proof without using the chain rule, use the definition of the derivative. We can prove something slightly more general. If
$$g(x) = \frac{d}{dx}f(x) = \lim_{h \to 0}\frac{f(x+h) - f(x)}{h}$$
and $a(x) = f(cx)$ where $c$ is any nonzero real number,
\begin{align} \frac{d}{dx} a(x) &= \lim_{h \to 0} \frac{a(x+h) - a(x)}{h} \\ &= \lim_{h \to 0} \frac{f(c(x+h)) - f(cx)}{h} \\ \end{align}
Letting $k = ch$, we note that $h \to 0$ if and only if $k \to 0$, so the above is equivalent to
\begin{align} \frac{d}{dx} a(x) &= \lim_{k \to 0} \frac{f(cx + k) - f(cx)}{k/c} \\ &= c \lim_{k \to 0} \frac{f(cx + k) - f(cx)}{k} \\ &= c g(cx) \\ \end{align}
Your case follows by setting $c = -1$.