1
$\begingroup$

I have function defined as $f(x,y)=\sqrt[3]{x^2+y}\cdot \ln{(x^2+y^2)}, f(0,0)=0$. I am evaluating derivatives:

$\frac{\partial f}{\partial x}(x,y)=\frac{1}{3}\cdot\frac{2x\ln{(x^2+y^2)}}{\sqrt[3]{(x^2+y)^2}}+\frac{2x\sqrt[3]{x^2+y}}{x^2+y^2}$ and $\frac{\partial f}{\partial y}(x,y)=\frac{1}{3}\frac{\ln{(x^2+y^2)}}{\sqrt[3]{(x^2+y)^2}} + \frac{2y\sqrt[3]{x^2+y}}{x^2+y^2}$.

I hope I haven't made any mistakes, however what troubles me is $[0,0]$, in this point function is defined as $f(0,0)=0$. How do I treat it(Do I have to differentiate it as 'different' function)?

Lastly, I am trying to determine what happens when $y=-x^2$, I tried to find out by using limits:

Let $y_o=-x_0^2$, then we have $\frac{\partial f}{\partial x}(x_0,y_0)=\lim_{x \rightarrow x_0}2x(\frac{\ln{(x^2+y_0^2)}}{3\sqrt[3]{(x^2+y_0)^2}}+\frac{\sqrt[3]{x^2+y_0}}{x^2+y_0^2})=\lim_{x \rightarrow x_0}2x(\frac{\ln{(x^2+x_0^4)}}{3\sqrt[3]{(x^2-x_0^2)^2}}+\frac{\sqrt[3]{x^2-x_0^2}}{x^2+x_0^4}) $

I can't get rid of $x^2-x_0^2$ in denominator, can you please help me with this problem? I am solving this kind of problems on my own for the first time, and also please don't use Taylor series, which I don't know. Thank you.

$\endgroup$
0
$\begingroup$

To compute partial derivatives at a ceratin point we can use the definition of partial derivatives.For example, $$f_x(0,0)=\lim_{h\rightarrow 0}\frac{f(h,0)-f(0,0)}{h}=\lim_{h\rightarrow 0}\frac{h^{2/3}\ln (h^2)}{h}=\lim_{h\rightarrow 0}\frac{\ln h^2}{h^{1/3}}.$$ This limit does not exists as $h\rightarrow 0$. (It approaches infinity or minus infinity). So at $(0,0)$ partial derivative $f_x$ is not defined at (0,0). You can check $f_y(0,0)$ in a similar way.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.