MCQ
If $u = {({x^2} + {y^2} + {z^2})^{3/2}}$, then ${\left( {{{\partial u} \over {\partial x}}} \right)^2} + {\left( {{{\partial u} \over {\partial y}}} \right)^2} + {\left( {{{\partial u} \over {\partial z}}} \right)^2} = $
  • A
    $9u$
  • $9{u^{4/3}}$
  • C
    $9{u^2}$
  • D
    ${u^{4/3}}$

Answer

Correct option: B.
$9{u^{4/3}}$
b
(b) $\frac{{\partial u}}{{\partial x}} = \frac{3}{2}{({x^2} + {y^2} + {z^2})^{1/2}}.2x$

$\therefore $ ${\left( {\frac{{\partial u}}{{\partial x}}} \right)^2} = \frac{9}{4}({x^2} + {y^2} + {z^2})4{x^2} =  9{x^2}({x^2} + {y^2} + {z^2})$

$\therefore $ ${\left( {\frac{{\partial u}}{{\partial x}}} \right)^2} + {\left( {\frac{{\partial u}}{{\partial y}}} \right)^2} + {\left( {\frac{{\partial u}}{{\partial z}}} \right)^2}$

$= 9\,({x^2} + {y^2} + {z^2})\,({x^2} + {y^2} + {z^2})$

$= 9\,{({x^2} + {y^2} + {z^2})^2}$ = $9.{u^{4/3}}$.

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