3
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
5
$\begingroup$

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:$$\lim_{h\to0}\frac{f\bigl(x+h(-v)\bigr)-f(x)}h=-\lim_{h\to0}\frac{f(x-hv)-f(x)}{-h}=-\nabla.$$

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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}

and

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

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.