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Math Help - Confusion about Total Derivative

  1. #1
    Senior Member slevvio's Avatar
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    Confusion about Total Derivative

    Hello everyone, I was a little confused about something and hope that someone can help me.

    Let  U \subseteq \mathbb{R} be open, and if a function  f : U \rightarrow \mathbb{R} is differentiable on  U, then for each u\in U there must exist a linear map Df(u):\mathbb{R} \rightarrow \mathbb{R} such that

    \displaystyle\lim_{h\rightarrow 0} \displaystyle\frac{| f(u+h)-f(u) - Df(u)h|}{|h|} = 0

    Then this is true if and only if Df(u) = f'(u) = \lim_{h \rightarrow 0} \displaystyle\frac{f(u+h)-f(u)}{h}. So doesn't this mean that if we can differentiate f, it's then differentiable as a linear map? I know that sounds stupid, but I'm trying to think of the result that if the partial derivatives of f are continuous, then f is differentiable in this special case. So is continuity of the derivative not important in the real case?

    What about parametric equations, for example f: \mathbb{R} \rightarrow \mathbb{R}^2 where f(x) = (y_1(x), y_2(x)). On wikipedia it just says that the derivative of f is f'(x) =   (y_1'(x), y_2'(x)). But do the derivatives of y_1, y_2 only have to exist for f to be differentiable? Does it matter if they're not continuous?

    Thanks for any help
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  2. #2
    Senior Member slevvio's Avatar
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    Oct 2007
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    Re: Confusion about Total Derivative

    Sorry, I've discovered that this is a known result:

    If \gamma:\mathbb{R} \rightarrow \mathbb{R}^m is differentiable if and only if its component functions are differentiable in the sense of single-variable calculus.
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