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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)

I had two questions:

  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.

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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.

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  • $\begingroup$ Thank you José Carlos. Just so I am clear, is it correct to say, in general, that $\lim\limits_{(-h) \to 0} \frac{f(x+(-h)a) - f(x)}{(-h)}$ = $\lim\limits_{(h) \to 0} \frac{f(x+ha) - f(x)}{(h)}$? $\endgroup$ – Cola Nov 10 '18 at 23:21
  • $\begingroup$ Yes, that is correct. $\endgroup$ – José Carlos Santos Nov 11 '18 at 8:13

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