I am a little bit confused by the various theorems concerning the differentiability of a multivariable function.
Let $f : D \subseteq R^n \to R$ have all directional derivatives in point $p$. Does it directly imply that $f$ is differentiable in $p$? I know that the opposite is true: If $f$ were differentiable, it would imply that it has directional derivatives.
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José Carlos Santos
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Aemilius
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No. For $f$ to be differentiable you need that all the partial derivatives exist AND are continous. The converse is also true. – Dog_69 Jun 09 '18 at 07:57
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2@Dog_69 You are wrong. If $f\colon\mathbb{R}\longrightarrow\mathbb{R}$ is defined by $f(x)=x^2\sin\left(\frac1x\right)$ if $x\neq0$ and $f(0)=0$, then $f'$ is discontinuous at $0$. However, $f$ is differentiable there. – José Carlos Santos Jun 09 '18 at 08:00
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@JoséCarlosSantos, why is $f$ differentiable in $0$? I am pretty sure that the definition of differentiability requires continuous derivatives. – Aemilius Jun 09 '18 at 08:56
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1The function $f$ is differentiable at $0$ because the limit $\lim_{x\to0}\frac{f(x)-f(0)}{x}$ exists (it is equal to $0$). And, no, being differentiable does not require continuous derivatives. – José Carlos Santos Jun 09 '18 at 09:16
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@JoséCarlosSantos: Yes, you are right.Thanks. If all the partial derivatives of $f$ exist AND are continuous then $f$ is defferentiable. BUT, there are differentiable functions which do not have continuous partial derivatives. – Dog_69 Jun 09 '18 at 10:34
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No, this is not true. Take, for instance$$\begin{array}{rccc}f\colon&\mathbb{R}^2&\longrightarrow&\mathbb{R}\\&(x,y)&\mapsto&\begin{cases}\frac{x^2y}{x^4+y^2}&\text{ if }(x,y)\neq(0,0)\\0&\text{ otherwise.}\end{cases}\end{array}$$You can check that, at $(0,0)$, every directional derivative exists. However, $f$ is not differentiable there.
José Carlos Santos
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The directional derivatives all exist at the origin (which is what was required), but they are not all zero. – Hans Lundmark Jun 03 '22 at 11:30
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1$$f_x(0,0)=\lim_{h\to0}\frac{f(h,0)-f(0,0)}h=\lim_{h\to0}\frac0h=\lim_{h\to0}0=0.$$ – José Carlos Santos Feb 01 '23 at 19:42
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Ler $$f(x,y)=\begin{cases} 1 & x^2< y < 2x^2\\ 0 & {\rm otherwise}\end{cases}$$ The directional derivatives are equal $0.$ The function is not continuous at $(0,0).$
Ryszard Szwarc
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