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Math Help - Directional derivative

  1. #1
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    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}

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  2. #2
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    What is your definition of
    1) Differentiable

    2) Differentiable in the direrction v?
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  3. #3
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    Quote Originally Posted by HallsofIvy View Post
    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<br />
    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.
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  4. #4
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    Quote Originally Posted by bram kierkels View Post
    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<br />
    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?
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