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how do I find the partial deritative of $$ f(x,y) = \sqrt[3]{x^3+y^3}$$

If I use normal rules I get $f_x = x^2(x^3+y^3)^{-\frac{2}{3}}$

so $f_x(0,0) = 0$

But if I calculate by definition I get

$$f_x(0,0) = \lim_{h\to{0}} \frac{f(0+h,0) - f(0,0)}{h} = 1$$

why does this happpen?

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  • $\begingroup$ your evaluation for $f_x(0,0)$ is wrong $\endgroup$
    – janmarqz
    Commented Aug 12, 2015 at 19:31

2 Answers 2

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Technically speaking, $f_x (0,0) \neq 0$ rather than $\frac{0}{0}$. More precisely, $(0, 0)$ is a singularity of $\frac{\partial f}{\partial x}$, so the definition is not expected to work there.

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The calculation $f_x = x^2(x^3 + y^3)^{-\frac{2}{3}}$ is valid for $(x,y)\neq (0,0)$, but not for $(x,y) = (0,0)$ as $(x^3 + y^3)^{-2/3}$ is not defined at $(0,0)$.

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