We are given a coordinate $x$ and a function $f(x)$ at each point on $x$. We need to compute the derivative at a point $x$. For this, we usually choose two points $x+dx$ and $x$ (first principles). Can we also choose the points $x-dx$ and $x$ and get the same derivative?

If not, why?

If yes, why do we get opposite values for directional derivatives in two opposite directions?

  • $\begingroup$ Technically $dx$ is not a number, and derivative is NOT $$\frac{f(x+dx)-f(x)}{dx}$$ which doesn't quite make sense. $\endgroup$ – Vim Sep 6 '17 at 10:47

The standad way of defining $f'(x)$ is$$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}h.$$If you want to use $-h$ instead of $h$, you can. But you will have to to do it in a consistent way. That is,$$f'(x)=\lim_{h\to0}\frac{f(x-h)-f(x)}{-h}.$$

If you are interested in directional derivatives, the situation is somewhat different. In this case we have a scalar function $f$ of $n$ variables and a vector $v\in\mathbb{R}^n$. Then the directional derivative of $f$ at $x$ with respect to $v$ is$$\lim_{h\to0}\frac{f(x+hv)-f(x)}h.$$This measures how fast $f$ grows near $x$ in the direction given by $v$. And what happens if $v$ gets replaced by $-v$? Well, if the growth in one direction is $\nabla$, then the growth in the opposite direction is $-\nabla$. That's is easy to prove:$$\lim_{h\to0}\frac{f\bigl(x+h(-v)\bigr)-f(x)}h=-\lim_{h\to0}\frac{f(x-hv)-f(x)}{-h}=-\nabla.$$

  • $\begingroup$ Will both yeild same answer? $\endgroup$ – Joe Sep 6 '17 at 10:46
  • $\begingroup$ @Joe Yes, they will. $\endgroup$ – José Carlos Santos Sep 6 '17 at 10:47
  • $\begingroup$ This is what I thought. But then why is it that we get opposite values for directional derivatives in two opposite directions? $\endgroup$ – Joe Sep 6 '17 at 10:51
  • $\begingroup$ I mean in one dimension, directional derivative along positive $x$ direction is given by your first equation (in your answer) and along negative $x$ direction by your second equation. But they are equal and opposite $\endgroup$ – Joe Sep 6 '17 at 10:56
  • $\begingroup$ @Joe Because that's a different concept! Your question suggested that you were interested in functions of a single variable. I shall edit my answer. $\endgroup$ – José Carlos Santos Sep 6 '17 at 10:57

You are given "points" $x$ and $x+\Delta x$. Nowhere is it stated that $\Delta x$ is a positive number, though it seems to be assumed a lot of the time. More to the point, the limit, $\displaystyle \lim_{\Delta x \to 0}\frac{f(x+\Delta x) - f(x)}{\Delta x}$, makes no assumptions on the sign of $\Delta x$. The limit must exists as $\Delta x$ approaches $0$ from either side.

But you need to watch how the negative sign is used.

\begin{align} \lim_{\Delta x \to 0}\frac{f(x-\Delta x) - f(x)}{\Delta x} &= -\lim_{\Delta x \to 0}\frac{f(x+(-\Delta x)) - f(x)}{(-\Delta x)} \\ &= -f'(x) \end{align}


$$\lim_{(-\Delta x) \to 0}\frac{f(x+\Delta x) - f(x)}{\Delta x} = f'(x)$$


Consider a one-dimensional example, such as $f(x) = x^2$. It could locally represent be the one-dimensional restriction of your two-dimensional function, along the direction of interest. Then the derivative at a point $x_0$ is given by $$ f´(x_0) = lim_{\epsilon \to 0}\frac{f(x_0 + \epsilon) - f(x_0)}{\epsilon} = \\ lim_{\epsilon \to 0}\frac{(x_0 + \epsilon)^2 - x_0^2}{\epsilon} = 2x_0 $$ or alternatively, following your post $$ f´(x_0) = lim_{\epsilon \to 0}\frac{f(x_0 ) - f(x_0 - \epsilon)}{\epsilon} = \\ lim_{\epsilon \to 0}\frac{x_0 ^2 - (x_0-\epsilon)^2}{\epsilon} = 2x_0 $$ and you can check how the two minus sign cooperate, in order for sign of the derivative to be the same.


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