# Directional derivative

• Dec 2nd 2010, 04:08 AM
bram kierkels
Directional derivative
Let $E$ be a subset of $\mathbb{R}^n, f:E\rightarrow \mathbb{R}^m$ be a function, $x_0$ be an interior point of $E$, and let $v$ be a vector in $\mathbb{R}^n$. I want to show if $f$ is differentiable at $x_0$, then $f$ is also differentiable in the direction $v$ at $x_0$, and

$D_vf(x_0)=f'(x_0)v,\ \ \ \ \ \ \$

where $D_vf(x_0):=\lim_{t\rightarrow 0, t\geq 0}\frac{f(x_0)+tv)+f(x_0)}{t}$

Thanks
• Dec 2nd 2010, 04:14 AM
HallsofIvy
What is your definition of
1) Differentiable

2) Differentiable in the direrction v?
• Dec 2nd 2010, 05:43 AM
bram kierkels
Quote:

Originally Posted by HallsofIvy
What is your definition of
1) Differentiable

2) Differentiable in the direrction v?

I say $f$ is differentiable at $x_0$ with derivative $L$ if we have

$\lim_{x\rightarrow x_0,x\in E-{x_0}}\frac{||f(x)-(f(x_0)+L(x-x_0))||}{||x-x_0||}=0
$

where $||x||$ is the lenght of $x$ as measured in the $l^2$ metric

I say $f$ is differentiable in the direction $v$ at $x_0$ if the limit

$\lim_{t\rightarrow 0;t>o,x_0+tv\in E}\frac{f(x_0+tv)-f(x_0)}{t}$ exists.
• Dec 2nd 2010, 07:47 AM
Opalg
Quote:

Originally Posted by bram kierkels
I say $f$ is differentiable at $x_0$ with derivative $L$ if we have

$\lim_{x\rightarrow x_0,x\in E-{x_0}}\frac{||f(x)-(f(x_0)+L(x-x_0))||}{||x-x_0||}=0
$

where $||x||$ is the length of $x$ as measured in the $l^2$ metric

I say $f$ is differentiable in the direction $v$ at $x_0$ if the limit

$\lim_{t\rightarrow 0;t>o,x_0+tv\in E}\frac{f(x_0+tv)-f(x_0)}{t}$ exists.

So what happens if you put $x = x_0+tv$ in the first of those definitions?