# Directional derivatives and limit rules

I am having some problems understanding directional derivatives and limit rules. I was hoping someone might be able to help.

Basically, the result is the $$f'(x, -a)$$ = $$-f'(x,a)$$. To show this (after some manipulations, we have:

(1) $$f'(x, -a) = \lim\limits_{(-h) \to 0} \frac{f(x+(-h)a) - f(x)}{(-h)} = -f'(x,a)$$ (2)

1. Firstly, can I intepret the result as (1) the instantenous rate of change in the direction opposite to $$a$$ is equal to (2) the instantaneous rate of change in the opposite direction to $$a$$?

2. I don't understand the second equality. When I write out $$-f(x,a)$$ according to the definition, I have:

$$-f'(x, a) = \lim\limits_{h \to 0} \frac{-f(x+ ha) + f(x)}{h}$$

It's not obvious to me why the two are equal. Is it because $$\lim\limits_{h \to 0}$$ is equivalent to $$\lim\limits_{(-h) \to 0}$$?

Thank you.

The equality that you wish to prove means that the instantaneous rate of change in the opposite direction of $$a$$ is the symetric of the instantaneous rate of change in the direction of $$a$$. And the answer to your final question is affirmative.
• Thank you José Carlos. Just so I am clear, is it correct to say, in general, that $\lim\limits_{(-h) \to 0} \frac{f(x+(-h)a) - f(x)}{(-h)}$ = $\lim\limits_{(h) \to 0} \frac{f(x+ha) - f(x)}{(h)}$? – Cola Nov 10 '18 at 23:21