$\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 $ ${\left( {\frac{{\partial u}}{{\partial x}}} \right)^2} + {\left( {\frac{{\partial u}}{{\partial y}}} \right)^2} + {\left( {\frac{{\partial u}}{{\partial z}}} \right)^2}$
$= 9\,({x^2} + {y^2} + {z^2})\,({x^2} + {y^2} + {z^2})$
$= 9\,{({x^2} + {y^2} + {z^2})^2}$ = $9.{u^{4/3}}$.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
| $Face:$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ |
| $Probability:$ | $0.1$ | $0.32$ | $0.21$ | $0.15$ | $0.05$ | $0.17$ |
The die is tossed and you are told that either face $1$ or $2$ has turned up. Then the probability that it is face $1$, is
