If u = (x^2 + y^2 + z^2)^{3/2},then left( fra | vedclass

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(B) Given $u = (x^2 + y^2 + z^2)^{3/2}$.First,find the partial derivative with respect to $x$:$\frac{\partial u}{\partial x} = \frac{3}{2}(x^2 + y^2 +

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$9u^{4/3}$

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(B) Given $u = (x^2 + y^2 + z^2)^{3/2}$.First,find the partial derivative with respect to $x$:$\frac{\partial u}{\partial x} = \frac{3}{2}(x^2 + y^2 + z^2)^{1/2} \cdot (2x) = 3x(x^2 + y^2 + z^2)^{1/2}$.Squaring this,we get:$\left( \frac{\partial u}{\partial x} \right)^2 = 9x^2(x^2 + y^2 + z^2)$.Similarly,by symmetry:$\left( \frac{\partial u}{\partial y} \right)^2 = 9y^2(x^2 + y^2 + z^2)$ and $\left( \frac{\partial u}{\partial z} \right)^2 = 9z^2(x^2 + y^2 + z^2)$.Adding these three expressions:$\left( \frac{\partial u}{\partial x} \right)^2 + \left( \frac{\partial u}{\partial y} \right)^2 + \left( \frac{\partial u}{\partial z} \right)^2 = 9(x^2 + y^2 + z^2)(x^2 + y^2 + z^2 + z^2) = 9(x^2 + y^2 + z^2)^2$.Since $u = (x^2 + y^2 + z^2)^{3/2}$,we have $u^{2/3} = (x^2 + y^2 + z^2)$.Therefore,$(x^2 + y^2 + z^2)^2 = (u^{2/3})^2 = u^{4/3}$.Thus,the sum is $9u^{4/3}$.

If $u = (x^2 + y^2 + z^2)^{3/2}$,then $\left( \frac{\partial u}{\partial x} \right)^2 + \left( \frac{\partial u}{\partial y} \right)^2 + \left( \frac{\partial u}{\partial z} \right)^2 = $

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