If u = ( x^2 + y^2 + z^2 )^ 3/2 , then ( u x )^2 + ( u y )^2 + ( … - Vidyadip

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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