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$f(x,y)=(\sqrt[3]{x} + \sqrt[3]{y})^3$ and let $P=(0,0)$. the partial derivative at $P$ are $$\frac{\partial f}{\partial x}(x,y)=\lim_{h\to0}\frac{f(0+h, 0)-f(0,0)}{h}=\frac{(\sqrt[3]{h})^3}{h}=1$$ and $$\frac{\partial f}{\partial y}(x,y)=\lim_{h\to0}\frac{f(0, 0+h)-f(0,0)}{h}=\frac{(\sqrt[3]{h})^3}{h}=1$$ and thus both partial derivative are continuous at $P$ and $f$ is differentiable. But when I derivate directly I get $$\frac{\partial f}{\partial x}(x,y)=\frac{(\sqrt[3]{x} + \sqrt[3]{y})^2}{\sqrt[3]{x^2}}$$ and $$\frac{\partial f}{\partial y}(x,y)=\frac{(\sqrt[3]{x} + \sqrt[3]{y})^2}{\sqrt[3]{y^2}}$$ which are undefined for $P=(0, 0)$ and thus clearly not continuous at $P$. So I'm not sure which derivative is correct and whether $f$ is differentiable or not?!?

Anton Vrdoljak
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CforLinux
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    You say, "both partial derivatives are continuous at $(0,0)$," but you have only shown they are defined at $(0,0)$. You need to consider the limits of the partials as $(x,y)\to(0,0)$. Also, continuity of the partials is sufficient for differentiability, but not necessary. – saulspatz Jun 20 '20 at 14:40

2 Answers2

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For $f$ to be differentiable at $(0,0)$ there must be a linear form $ax+by$ such that $$f(x,y)-(ax+by)$$ tends to zero quicker than $\sqrt{x^2+y^2}$ as $(x,y)\to 0$, that is $$\frac{f(x,y)-(ax+by)}{\sqrt{x^2+y^2}}\to0.$$ The numbers $a$ and $b$ must be the partial derivatives of $f$ at $(0,0)$, that is $a=1$ and $b=1$ so that $$\frac{f(x,y)-(ax+by)}{\sqrt{x^2+y^2}} =\frac{f(x,y)-x-y}{\sqrt{x^2+y^2}} =3\frac{\sqrt[3]{x^2}\sqrt[3]y+\sqrt[3]{x}\sqrt[3]{y^2}}{\sqrt{x^2+y^2}} .$$ But if we take $(x,y)=(t,t)$ we get $$\frac{f(x,y)-(ax+by)}{\sqrt{x^2+y^2}} =\frac{6t}{\sqrt{2}|t|}=\pm3\sqrt2$$ which certainly does not tend to zero as $t\to0$.

Therefore $f$ is not differentiable at $(0,0)$.

Angina Seng
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If $f(x,y)$ were differentiable at $(0,0)$, its directional derivative in the $(a,b)$ direction at $(0,0)$ would have to be $$a{\partial f \over \partial x}(0,0) + b{\partial f \over \partial y}(0,0) = a + b$$ But this doesn't hold in most directions. Suppose you choose $(a,b) = ({1 \over \sqrt{2}}, {1 \over \sqrt{2}})$. Then $$f(ta, tb) = f\bigg({t \over \sqrt{2}}, {t \over \sqrt{2}}\bigg)$$ $$= \bigg(2\bigg({t \over \sqrt{2}}\bigg)^{1 \over 3}\bigg)^3$$ $$= {8 \over \sqrt{2}} t$$ Thus $$\lim_{t \rightarrow 0} {f(t,t) \over t} = {8 \over \sqrt{2}}$$ This does not equal $a + b = {2 \over \sqrt{2}} = \sqrt{2}$, so the function is not differentiable at the origin.

Zarrax
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