Do first order partial derivatives of this function exist at $(0,1)$?

Question

Do partial derivatives of $f(x,y) = e^{x+y}[x^{\frac{1}{3}}(y-1)^{\frac{1}{3}}+y^{\frac{1}{3}}(x-1)^{\frac{2}{3}}]$ with respect to $x$ and $y$ exist at (0,1)?

Solving directly for $f_x$ and $f_y$ is quite complicated, so I tried the limit definition:

$f_x(x,y)=\lim\limits_{h\to 0} \frac{f(x+h,y)-f(x,y)}{h}$ ;$f_y(x,y)=\lim\limits_{h\to 0} \frac{f(x,y+h)-f(x,y)}{h}$.

Plugging in $(0,1)$ yields terms like $(-1)^{\frac{2}{3}}$ and $(h-1)^{\frac{2}{3}}$ which could have three solutions (1 real, 2 complex). And I am stuck and don't know how to proceed.

Have I made a mistake somewhere? What would be the right way to tackle the problem?

Answer 1

partial derivatives of $f$ do exist at $(0,1)$: we have $f(h,1)=e^{1+h}(1-h)^{2/3}=e(1+h+o(h))(1-\frac 23 h+o(h))=e(1+\frac 13h+o(h))$ and $f(0,1+h)=e^{1+h}(1+h)^{1/3}=e(1+\frac 43h+o(h))$. However, $f$ is not diffferentiable at $(0,1)$, as $f(h,1-h)=e(h^{2/3}-1+h)$.

Answer 2

I think you are not wrong. This function is $f:\Bbb R \rightarrow \Bbb C$ function which is already a complex function, you would probably get some complex answers. Also, a function may have more than one answers as it maybe a multi-valued function.