Confusion about Total Derivative

**slevvio** · August 31st 2011, 05:02 AM

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

**slevvio** · August 31st 2011, 06:13 AM

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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