Question

solve this question ​

Answer 1

Given Question

If

$\rm :\longmapsto\:u = log( {x}^{3} + {y}^{3} - {x}^{2}y - {xy}^{2})$

Prove that,

$\rm :\longmapsto\:\dfrac{ {\partial }^{2}u}{ {\partial x}^{2} } + 2\dfrac{ {\partial }^{2} u}{\partial x\partial y} + \dfrac{ {\partial }^{2} u}{ {\partial y}^{2} } = \dfrac{4}{ {(x + y)}^{2} }$

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:u = log( {x}^{3} + {y}^{3} - {x}^{2}y - {xy}^{2})$

$\rm :\longmapsto\:u = log[({x}^{3} + {y}^{3}) - ({x}^{2}y + {xy}^{2})]$

$\rm :\longmapsto\:u = log[(x + y)( {x}^{2} - xy + {y}^{2}) - xy(x + y)]$

$\rm :\longmapsto\:u = log[(x + y)( {x}^{2} - xy + {y}^{2} - xy)]$

$\rm :\longmapsto\:u = log[(x + y)( {x}^{2} - 2xy + {y}^{2})]$

$\rm :\longmapsto\:u = log[(x + y)({x - y)}^{2}]$

$\rm\implies \:u = log(x + y) + 2log(x - y)$

On differentiating partially w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{\partial u}{\partial x} = \dfrac{1}{x + y} + \dfrac{2}{x - y}$

On differentiating partially w. r. t. x, we get

$\rm :\longmapsto\:\boxed{\tt{ \frac{ {\partial }^{2} u}{ {\partial x}^{2} } = \frac{ - 1}{ {(x + y)}^{2} } - \frac{2}{ {(x - y)}^{2} }}} - - (1)$

Again, Consider

$\rm :\longmapsto\: \:u = log(x + y) + 2log(x - y)$

On differentiating partially w. r. t. y, we get

$\rm :\longmapsto\:\dfrac{\partial u}{\partial y} = \dfrac{1}{x + y} - \dfrac{2}{x - y}$

On differentiating partially w. r. t. y, we get

$\rm :\longmapsto\:\boxed{\tt{ \frac{ {\partial }^{2} u}{ {\partial y}^{2} } = \frac{ - 1}{ {(x + y)}^{2} } - \frac{2}{ {(x - y)}^{2} }}} - - - (2)$

Now, we have

$\rm :\longmapsto\:\dfrac{\partial u}{\partial x} = \dfrac{1}{x + y} + \dfrac{2}{x - y}$

On differentiating partially w. r. t. y, we get

$\rm :\longmapsto\:\boxed{\tt{ \dfrac{ {\partial }^{2} u}{\partial x\partial y} = \dfrac{ - 1}{ {(x + y)}^{2} } + \dfrac{2}{ {(x- y)}^{2} }}} - - - (3)$

Now, Consider

$\rm :\longmapsto\:\dfrac{ {\partial }^{2}u}{ {\partial x}^{2} } + 2\dfrac{ {\partial }^{2} u}{\partial x\partial y} + \dfrac{ {\partial }^{2} u}{ {\partial y}^{2} }$

$\rm \:  =  \: - \dfrac{1}{ {(x + y)}^{2} } - \dfrac{2}{ {(x - y)}^{2} } - \dfrac{2}{ {(x + y)}^{2} } + \dfrac{4}{ {(x - y)}^{2} } - \dfrac{1}{ {(x + y)}^{2} } - \dfrac{2}{ {(x - y)}^{2} }$

$\rm \:  =  \: - \dfrac{4}{ {(x + y)}^{2} }$

Hence, Proved

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FORMULA USED

$\purple{\rm :\longmapsto\: \boxed{\tt{ {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2}}}}$

$\purple{\rm :\longmapsto\:\boxed{\tt{ log(xy) = logx \: + \: logy}}}$

$\purple{\rm :\longmapsto\:\boxed{\tt{ log {x}^{y} = y \: logx}}}$

$\purple{\rm :\longmapsto\:\boxed{\tt{ \dfrac{d}{dx}logx = \frac{1}{x}}}}$

$\purple{\rm :\longmapsto\:\boxed{\tt{ \dfrac{d}{dx}x = 1}}}$

$\purple{\rm :\longmapsto\:\boxed{\tt{ \dfrac{d}{dx}k = 0}}}$

$\purple{\rm :\longmapsto\:\boxed{\tt{ \dfrac{d}{dx} \frac{1}{x} = \frac{ - 1}{ {x}^{2} } }}}$

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LEARN MORE

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$