Question

let u =log(x^2+y^2+z^2),prove that (d^2u/dx^2)+(d^2u/dy^2)(d^2u/dz^2)=2/(x^2+y^2+z^2)

Answer 1

Step-by-step explanation:

$\sf \: u = log( {x}^{2} + {y}^{2} + {z}^{2} )$

$\sf \: \frac{ \partial u}{ \partial x} = \frac{1}{ {x}^{2} + {y}^{2} + {z}^{2} } (2x + 0 + 0)$

$\sf \: \frac{ \partial u}{ \partial x} = \frac{2x}{ {x}^{2} + {y}^{2} + {z}^{2} }$

$\sf \: \frac{ \partial ^{2} u}{ \partial ^{2} x} = 2\frac{1( {x }^{2} + {y}^{2} + {z}^{2} ) - x(2x) }{ ({x}^{2} + {y}^{2} + {z}^{2} ) {}^{2} }$

$\sf \: \frac{ \partial ^{2} u}{ \partial ^{2} x} = 2\frac{{ - x }^{2} + {y}^{2} + {z}^{2} }{ ({x}^{2} + {y}^{2} + {z}^{2} ) {}^{2} } \: \: ....(i)$

$\large \sf \: \green{ \underline{ \sf \: SIMILARLY : - }}$

$\sf \: \frac{ \partial ^{2} u}{ \partial ^{2} y} = 2\frac{{ x }^{2} - {y}^{2} + {z}^{2} }{ ({x}^{2} + {y}^{2} + {z}^{2} ) {}^{2} } \: \: \: \: \: ......(ii)$

$\sf \: \frac{ \partial ^{2} u}{ \partial ^{2} z} = 2\frac{{ x }^{2} + {y}^{2} - {z}^{2} }{ ({x}^{2} + {y}^{2} + {z}^{2} ) {}^{2} } \: \: \: \: .....(iii)$

$\sf \: \frac{ \partial ^{2} u}{ \partial ^{2} x} + \frac{ \partial ^{2} u}{ \partial ^{2} y} + \frac{ \partial ^{2} u}{ \partial ^{2} z} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \sf \: = 2\frac{{ - x }^{2} + {y}^{2} + {z}^{2} }{ ({x}^{2} + {y}^{2} + {z}^{2} ) {}^{2} } + 2\frac{{ x }^{2} - {y}^{2} + {z}^{2} }{ ({x}^{2} + {y}^{2} + {z}^{2} ) {}^{2} } \sf \: + 2\frac{{ x }^{2} + {y}^{2} - {z}^{2} }{ ({x}^{2} + {y}^{2} + {z}^{2} ) {}^{2} }$

$\sf \: = 2 \bigg(\frac{{ - x }^{2} + {y}^{2} + {z}^{2} + {x}^{2} - {y }^{2} + {z}^{2} + {x}^{2} + {y}^{2} - {z}^{2} }{ ({x}^{2} + {y}^{2} + {z}^{2} ) {}^{2} } \bigg)$

$\sf \: = 2 \frac{ {x}^{2} + {y}^{2} + {z}^{2} }{( {x}^{2} + {y}^{2} + {z}^{2} ) {}^{2} }$

$\sf = \frac{2}{( {x}^{2} + {y}^{2} + {z}^{2} )}$

$\sf \large{ \green{ \sf \: proved}}$