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Let $U\subseteq \Bbb R^n$ be open.

If $f$ has continuous first partial derivatives in $U$ then $f$ is differentiable in $U$.


Let's prove that if $f$ is differentiable at $a$ then $T(h)=Df(a).h$ ,where T(h) is total derivative.

Let's set $T(h)= (\alpha_1.h+\alpha_2.h+...+\alpha_n.h)=(\alpha_1,...,\alpha_n).h $

Since $f$ has continuous derivatives, $$\lim_{h\to0}\frac{f(a+h)-f(a)-T(h)}{\Vert h\Vert}=0 \tag 1$$ by the first order approximation theorem.

Set $h:=te_i$ for $t>0$

$$\begin{align} \lim_{h\to0}\frac{f(a+h)-f(a)-T(h)}{\Vert h\Vert}&=\lim_{h\to0}\frac{f(a+te_i)-f(a)-\alpha_it}{t} \\ &=\frac{f(a+te_i)-f(a)}{t}-\alpha_i \end{align}$$

$\lim_{t\to 0^+}\frac{f(a+te_i)-f(a)}{t}=\alpha_i$ by the equation (1)

We also need to check this limit for $t<0$

Do I prove the theorem? Is this proof enough? Do I have mistakes ? Please can one correct my proof? Thank you:)

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Please define continuous differentiability. – Git Gud Jul 13 '13 at 20:38
I imagine they mean "has continuous first partial derivatives". But what is the first order approximation theorem? Is $Df$ what I'm accustomed to seeing written $\nabla f$? Are the periods what I'm accustomed to seeing written as $\cdot$? – dfeuer Jul 13 '13 at 20:40
B11b, it's okay that you're (apparently) not a native English speaker, but that means you have to be extra careful when explaining things. In an online mathematical forum, where being clear is more important than sounding good, it is best to use short sentences with simple structure and explain even the bits that seem obvious. – dfeuer Jul 13 '13 at 20:49
@dfeuer by the way, yes, I speak English only for academic field. Everthing I have written is so clear and understandable. Only you make such a comment to me. It is interesting! – B11b Jul 13 '13 at 21:06
@GitGud Is the proof enough? What is your comment instead of the definiton of continuously differentiable ? Thank you. – B11b Jul 13 '13 at 21:12

Hint: for $x=(x_1,\ldots,x_n), y=(y_1,\ldots,y_n),$ write $$f(x)-f(y) =$$ $$ f(x_1,\ldots,x_n) - f(y_1,x_2,\ldots,x_n) + f(y_1,x_2,\ldots,x_n) - f(y_1,y_2,x_3,\ldots,x_n) + f(y_1,y_2,x_3,\ldots,x_n) - \ldots - f(y_1,\ldots,y_n)$$ and apply the mean value theorem multiple times.

You will find $$ f(x)-f(y) - \nabla f(y)(x-y) = \sum_{i=1}^n \left(\partial_if(z_i)-\partial_if(y)\right)(x_i-y_i). $$ Do you see how to finish off the proof? Extra bonus: to show differentiability at an interior point $y$, it is actually enough to assume $n-1$ of the $\partial_i f$ exist in a ball around $y$ and are continuous at $y$, and that the $n$th partial just exists at $y$.

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Am I correct in thinking that it's also true that if the first partials exist in an open set and are continuous at a particular point, then the function is differentiable at that point? – dfeuer Jul 13 '13 at 21:34
Yes you are correct, that is what my proof above is leading you towards. – nullUser Jul 13 '13 at 21:42
I misread what you wrote. What you said was already strictly more general than what I was thinking. – dfeuer Jul 13 '13 at 21:56

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