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I am required to find $\dfrac{\partial f}{\partial x}(\textbf{a})$ where $f(x,y) = \sqrt[5]{(3x^5+2y^5)}$ at $\textbf{a} = (0,0)$. I have evaluated $\dfrac{\partial f}{\partial x}(x,y) = \dfrac{3x^4}{\sqrt[5]{(3x^5 + 2y^5)^4}}$. However, I believe at $(0,0)$ this does not exist. However I have been informed that there does indeed exist a value. Where have I gone wrong?

LogicAndTruth
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    You need to use the definition of partial derivative. So $\frac{\partial f}{\partial x}_{(0,0)} = \frac{d}{dx} (f(x,0))$ – Anon Mar 31 '21 at 16:51

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You need to go back to the definition of the partial derivative as a limit. You would have \begin{equation*} \frac{\partial f}{\partial x}(0,0)=\lim_{h\to 0}\frac{f(h,0)-f(0,0)}{h}=\lim_{h\to 0}\frac{(3h^{5})^{\frac{1}{5}}}{h}=3^{\frac{1}{5}}\end{equation*} in particular for this situation.

JWP_HTX
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