# Thread: Confusion about Total Derivative

1. ## Confusion about Total Derivative

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

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

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

Then this is true if and only if $\displaystyle Df(u) = f'(u) = \lim_{h \rightarrow 0} \displaystyle\frac{f(u+h)-f(u)}{h}$. 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 $\displaystyle f: \mathbb{R} \rightarrow \mathbb{R}^2$ where $\displaystyle f(x) = (y_1(x), y_2(x))$. On wikipedia it just says that the derivative of f is $\displaystyle f'(x) = (y_1'(x), y_2'(x)).$ But do the derivatives of $\displaystyle y_1, y_2$ only have to exist for f to be differentiable? Does it matter if they're not continuous?

Thanks for any help

2. ## Re: Confusion about Total Derivative

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

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