Question

If given that ;

$\bf {e}^{xyz} = u$

Then Prove that ;

${ \bigstar { \underline { \boxed { \bf { \blue { \dfrac{\partial³}{\partial x . \partial y . \partial z } = ( 1 + 3xyz + x²y²z² ) u }}}}}}{\bigstar}$

​

Answer 1

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

Given that

$\rm :\longmapsto\:u = {e}^{xyz}$

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

$\rm :\longmapsto\:\dfrac{\partial }{\partial x}u = \dfrac{\partial }{\partial x}{e}^{xyz} \:$

We know

$\boxed{ \tt{ \: \dfrac{d}{dx} {e}^{x} = {e}^{x} \: }}$

So, using this, we get

$\rm :\longmapsto\:\dfrac{\partial }{\partial x}u = {e}^{xyz} \dfrac{\partial }{\partial x}xyz \:$

$\rm :\longmapsto\:\dfrac{\partial u}{\partial x} = {e}^{xyz} yz \:$

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

$\rm :\longmapsto\:\dfrac{\partial }{\partial y}\dfrac{\partial u}{\partial x} = \dfrac{\partial }{\partial y}{e}^{xyz} yz \:$

$\rm :\longmapsto\:\dfrac{ {\partial }^{2}u}{\partial x {\partial y} } = yz\dfrac{\partial }{\partial y}{e}^{xyz} + {e}^{xyz}\dfrac{\partial }{\partial y}yz$

$\rm :\longmapsto\:\dfrac{ {\partial }^{2}u}{\partial x {\partial y} } = yz{e}^{xyz}(xz) + {e}^{xyz}z$

$\rm :\longmapsto\:\dfrac{ {\partial }^{2}u}{\partial x {\partial y} } =xy {z}^{2} {e}^{xyz} + {e}^{xyz}z$

$\rm :\longmapsto\:\dfrac{ {\partial }^{2}u}{\partial x {\partial y} } =(xy {z}^{2} + z){e}^{xyz}$

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

$\rm :\longmapsto\:\dfrac{ {\partial }^{3}u}{\partial x {\partial y} \partial z} =\dfrac{\partial }{\partial z}(xy {z}^{2} + z){e}^{xyz}$

$\rm :\longmapsto\:\dfrac{ {\partial }^{3}u}{\partial x {\partial y} \partial z} ={e}^{xyz}\dfrac{\partial }{\partial z}(xy {z}^{2} + z) + (xy {z}^{2} + z)\dfrac{\partial }{\partial z}{e}^{xyz}$

$\rm :\longmapsto\:\dfrac{ {\partial }^{3}u}{\partial x {\partial y} \partial z} ={e}^{xyz}(2xyz + 1) + (xy {z}^{2} + z){e}^{xyz}(xy)$

$\rm :\longmapsto\:\dfrac{ {\partial }^{3}u}{\partial x {\partial y} \partial z} ={e}^{xyz}(2xyz + 1) + ( {x}^{2} {y}^{2} {z}^{2} + xyz){e}^{xyz}$

$\rm :\longmapsto\:\dfrac{ {\partial }^{3}u}{\partial x {\partial y} \partial z} =[2xyz + 1 + {x}^{2} {y}^{2} {z}^{2} + xyz]{e}^{xyz}$

$\rm :\longmapsto\:\dfrac{ {\partial }^{3}u}{\partial x {\partial y} \partial z} =\bigg[ 1 + 3xyz + {x}^{2} {y}^{2} {z}^{2}\bigg]{e}^{xyz}$

$\rm :\longmapsto\:\dfrac{ {\partial }^{3}u}{\partial x {\partial y} \partial z} =\bigg[ 1 + 3xyz + {x}^{2} {y}^{2} {z}^{2}\bigg]u$

Hence,

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: \dfrac{ {\partial }^{3}u}{\partial x {\partial y} \partial z} =\bigg[ 1 + 3xyz + {x}^{2} {y}^{2} {z}^{2}\bigg]u \: }}}$

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Additional Information :-

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: \frac{ {\partial }^{2}u}{\partial x\partial y} = \frac{ {\partial }^{2} u}{\partial y\partial x} \: }}}$

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