Differentiating f(-x): Proving Correctness

[tade]

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?

 

[phyzguy]

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

 

[tade]

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

What are the specifics?

 

[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.

 

[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 $c = -1$.