1
$\begingroup$

One of the definitions for a directional derivative of $f$ in direction $v$ is as follows: $$ \nabla_v f(x) = \lim_{h \rightarrow 0}\frac{f(x+hv) - f(x)}{h\lvert v \rvert} $$

According to wikipedia, if $f$ is differentiable at $x$, $$ \nabla_v f(x) = \nabla f(x) \cdot \frac{v}{\lvert v \rvert}. $$

Why is this the case?

$\endgroup$
1
  • 1
    $\begingroup$ On this site, you're recommended to type out the equations instead of posting links $\endgroup$
    – Alan Chung
    Jun 15, 2023 at 18:10

1 Answer 1

0
$\begingroup$

If $f$ is differentiable at $x$ then this means that there is a vector $\nabla f(x)$ such that $$f(x + u) = f(x) + u\cdot \nabla f(x) + o(|u|)$$ (and this defines what is meant by $\nabla f(x)$). Then $$\frac{f(x + hv) - f(x)}{h|v|} = \frac{v\cdot \nabla f(x)}{|v|} + o(1)$$ and the result follows.

$\endgroup$
1
  • $\begingroup$ got it, thanks! $\endgroup$ Jun 15, 2023 at 18:52

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .