Question Number 53378 by Necxx last updated on 21/Jan/19
$${if}\:{u}={e}^{{xyz}} \:{then}\:{u}_{{xyx}} =? \\ $$$$\left.{a}\left.\right){u}\left(\left({xyz}\right)^{\mathrm{2}} +\mathrm{3}{xyz}+\mathrm{1}\right)\:{b}\right){u}\left(\mathrm{3}\left({xyz}\right)^{\mathrm{2}} +\mathrm{1}\right) \\ $$$$\left.{c}\right){u}\left(\left({xyz}\right)^{\mathrm{2}} +\mathrm{2}{yz}+\mathrm{1}\right) \\ $$$$ \\ $$$${please}\:{help} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 21/Jan/19
![u=e^(xyz) u_x =e^(xyz) ×(∂/∂x)(xyz)=yze^(xyx) u_(xy) =(∂/∂y)(yze^(xyz) )=e^(xyz) ×z(∂/∂y)(y)+yz×(∂/∂y)(e^(xyz) ) =ze^(xyz) +yz×e^(xyz) ×xz =e^(xyz) (z+xyz^2 ) u_(xyz) =e^(xyz) ×(∂/∂z)(z+xyz^2 )+(z+xyz^2 )×(∂/∂z)(e^(xyz) ) =e^(xyz) ×(1+2xyz)+(z+xyz^2 )×e^(xyz) (xy) =u[1+2xyz+xyz+x^2 y^2 z^2 ] =u[1+3xyz+x^2 y^2 z^2 ] so option a is correct](https://www.tinkutara.com/question/Q53387.png)
$${u}={e}^{{xyz}} \\ $$$${u}_{{x}} ={e}^{{xyz}} ×\frac{\partial}{\partial{x}}\left({xyz}\right)={yze}^{{xyx}} \\ $$$${u}_{{xy}} =\frac{\partial}{\partial{y}}\left({yze}^{{xyz}} \right)={e}^{{xyz}} ×{z}\frac{\partial}{\partial{y}}\left({y}\right)+{yz}×\frac{\partial}{\partial{y}}\left({e}^{{xyz}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={ze}^{{xyz}} +{yz}×{e}^{{xyz}} ×{xz} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={e}^{{xyz}} \left({z}+{xyz}^{\mathrm{2}} \right)\:\:\:\:\:\:\:\:\: \\ $$$${u}_{{xyz}} ={e}^{{xyz}} ×\frac{\partial}{\partial{z}}\left({z}+{xyz}^{\mathrm{2}} \right)+\left({z}+{xyz}^{\mathrm{2}} \right)×\frac{\partial}{\partial{z}}\left({e}^{{xyz}} \right) \\ $$$$\:\:\:={e}^{{xyz}} ×\left(\mathrm{1}+\mathrm{2}{xyz}\right)+\left({z}+{xyz}^{\mathrm{2}} \right)×{e}^{{xyz}} \left({xy}\right) \\ $$$$={u}\left[\mathrm{1}+\mathrm{2}{xyz}+{xyz}+{x}^{\mathrm{2}} {y}^{\mathrm{2}} {z}^{\mathrm{2}} \right] \\ $$$$={u}\left[\mathrm{1}+\mathrm{3}{xyz}+{x}^{\mathrm{2}} {y}^{\mathrm{2}} {z}^{\mathrm{2}} \right]\:\:{so}\:{option}\:{a}\:{is}\:{correct} \\ $$
Commented by Necxx last updated on 21/Jan/19
$${exactly}…..\:{Thanks}\:{so}\:{much} \\ $$