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STD 12 - 5. continuity and differentiation — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 5. continuity and differentiation4 Marks
MCQ
If $u = \log ({x^2} + {y^2}),$ then ${{{\partial ^2}u} \over {\partial {x^2}}} + {{{\partial ^2}u} \over {\partial {y^2}}} = $
  • A
    ${1 \over {{x^2} + {y^2}}}$
  • ✓
    $0$
  • C
    ${{{x^2} - {y^2}} \over {{{({x^2} + {y^2})}^2}}}$
  • D
    ${{{y^2} - {x^2}} \over {{{({x^2} + {y^2})}^2}}}$
✓

Answer

Correct option: B.
$0$
b
(b) $u = \log ({x^2} + {y^2})\,$

$\therefore \frac{{\partial u}}{{\partial x}} = \frac{1}{{{x^2} + {y^2}}}.2x$

$\frac{{{\partial ^2}u}}{{\partial {x^2}}} = \frac{{({x^2} + {y^2}).2 - 2x.2x}}{{{{({x^2} + {y^2})}^2}}}$$ = \frac{{2({y^2} - {x^2})}}{{{{({x^2} + {y^2})}^2}}}$’

$\frac{{\partial u}}{{\partial y}} = \frac{1}{{{x^2} + {y^2}}}.2y$

$\therefore \frac{{{\partial ^2}u}}{{\partial {y^2}}} = \frac{{({x^2} + {y^2})\,.\,2 - 2y.2y}}{{{{({x^2} + {y^2})}^2}}} = \frac{{2({x^2} - {y^2})}}{{{{({x^2} + {y^2})}^2}}}$

$\therefore $ $\frac{{{\partial ^2}u}}{{\partial {x^2}}} + \frac{{{\partial ^2}u}}{{\partial {y^2}}} = 0$.

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