Question Number 71729 by peter frank last updated on 19/Oct/19
$${find}\:{dU}\:\:\:{if}\:\:\:{U}={x}^{\mathrm{2}} {e}^{\frac{{x}}{{y}}} \\ $$$$ \\ $$
Commented by peter frank last updated on 19/Oct/19
$${thank}\:{you} \\ $$
Commented by Prithwish sen last updated on 19/Oct/19
![ylnu = 2yln(x) +x du = [2xe^(x/y) dx+((x^2 e^(x/y) )/y) dx− (x^3 /y^2 )e^(x/y) dy] please check](Q71733.png)
$$\mathrm{ylnu}\:=\:\mathrm{2yln}\left(\mathrm{x}\right)\:+\mathrm{x} \\ $$$$\boldsymbol{\mathrm{du}}\:=\:\left[\mathrm{2}\boldsymbol{\mathrm{xe}}^{\frac{\boldsymbol{\mathrm{x}}}{\boldsymbol{\mathrm{y}}}} \boldsymbol{\mathrm{dx}}+\frac{\boldsymbol{\mathrm{x}}^{\mathrm{2}} \boldsymbol{\mathrm{e}}^{\frac{\boldsymbol{\mathrm{x}}}{\boldsymbol{\mathrm{y}}}} }{\boldsymbol{\mathrm{y}}}\:\boldsymbol{\mathrm{dx}}−\:\frac{\boldsymbol{\mathrm{x}}^{\mathrm{3}} }{\boldsymbol{\mathrm{y}}^{\mathrm{2}} }\boldsymbol{\mathrm{e}}^{\frac{\boldsymbol{\mathrm{x}}}{\boldsymbol{\mathrm{y}}}} \:\boldsymbol{\mathrm{dy}}\right]\:\:\:\boldsymbol{\mathrm{please}}\:\boldsymbol{\mathrm{check}} \\ $$
Answered by mind is power last updated on 19/Oct/19
$$\mathrm{du}=\frac{\partial\mathrm{u}}{\partial\mathrm{x}}\mathrm{dx}+\frac{\partial\mathrm{u}}{\partial\mathrm{y}}\mathrm{dy} \\ $$$$\frac{\partial}{\partial\mathrm{x}}\left(\mathrm{x}^{\mathrm{2}} \mathrm{e}^{\frac{\mathrm{x}}{\mathrm{y}}} \right)=\left(\mathrm{2x}+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{y}}\right)\mathrm{e}^{\frac{\mathrm{x}}{\mathrm{y}}} \\ $$$$\frac{\partial}{\partial\mathrm{y}}\left(\mathrm{x}^{\mathrm{2}} \mathrm{e}^{\frac{\mathrm{x}}{\mathrm{y}}} \right)=−\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{y}^{\mathrm{2}} }\mathrm{e}^{\frac{\mathrm{x}}{\mathrm{y}}} \\ $$$$\Rightarrow\mathrm{dU}=\left(\mathrm{2x}+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{y}}\right)\mathrm{e}^{\frac{\mathrm{x}}{\mathrm{y}}} \mathrm{dx}−\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{y}^{\mathrm{2}} }\mathrm{e}^{\frac{\mathrm{x}}{\mathrm{y}}} \mathrm{dy} \\ $$
Commented by peter frank last updated on 19/Oct/19
$${thank}\:{you} \\ $$