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Math Help - Directional Derivative Proof

  1. #1
    Junior Member tedii's Avatar
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    Directional Derivative Proof

    Let A\subset\Re^m ; let  f:A\rightarrow\Re^n Show that if  f'(a;u) exists, then  f'(a:cu) exist and equals cf'(a;u).

    Some people in my class were trying to say that we know that by theorem

    f'(a:u)=Df(a)*u

    But I disagree since we don't know that Df(a) exist but I don't know how to prove this result. Any ideas would be helpful.
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  2. #2
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    Quote Originally Posted by tedii View Post
    Let A\subset\Re^m ; let  f:A\rightarrow\Re^n Show that if  f'(a;u) exists, then  f'(a:cu) exist and equals cf'(a;u).

    Some people in my class were trying to say that we know that by theorem

    f'(a:u)=Df(a)*u

    But I disagree since we don't know that Df(a) exist but I don't know how to prove this result. Any ideas would be helpful.
    Does f'(a;u) indicate the derivative of f in the direction of unit vector u? Yes, you are right. The derivative in a specific direction may exist even if the function is not "differentiable".

    You can prove it directly from the definition:

    The definition of the derivative, in the direction of vector u, is f'(a;u)= \lim_{h\to 0}\frac{f(a+ hu)- f(a)}{h}

    It follows that f'(a;cu)= \lim_{h\to 0}\frac{f(a+ hcu)- f(a)}{h}.

    Multiply numerator and denominator by c to get
    \lim_{h\to 0}c\frac{f(a+chu)- f(a)}{ch}

    Now let, k= ch. Since c is a constant, k goes to 0 as h goes to 0 and we can write that
    c\lim_{k\to 0}\frac{f(a+ku)- f(a)}{k}= c f'(a;u).
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  3. #3
    Junior Member tedii's Avatar
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    Tricky tricky, your right that makes perfect sense now, and I can wrap my head around that way of doing it. Thank you very much.
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