differentiation

Question: If $z=\frac{{{({{x}^{4}}+{{y}^{4}})}^{1/3}}}{{{({{x}^{3}}+{{y}^{3}})}^{1/4}}}$, then $x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=$

1) $\frac{1}{12}z$

2) $\frac{1}{4}z$

3) $\frac{1}{3}z$

4) $\frac{7}{12}z$

Solution:

Explanation: No Explanation

Partial differentiation

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