Question Number 207954 by solihin last updated on 01/Jun/24
$$\:\underline{\boldsymbol{{x}}} \\ $$
Commented by MathematicalUser2357 last updated on 02/Jun/24
$${I}\:{see}\:\underline{\boldsymbol{{x}}}\:{in}\:{tinku}\:{tara} \\ $$
Answered by AliJumaa last updated on 01/Jun/24
$${f}\left({x}_{\mathrm{1}} ,{x}_{\mathrm{2}} ,{y}\right)={yx}_{\mathrm{1}} +{yx}_{\mathrm{2}} −{x}_{\mathrm{1}} \\ $$$${df}\left({x}_{\mathrm{1}} ,{x}_{\mathrm{2}} ,{y}\right)=\frac{\partial}{\partial{x}_{\mathrm{1}} }{f}+\frac{\partial}{\partial{x}_{\mathrm{2}} }{f}+\frac{\partial}{\partial{y}}{f} \\ $$$$\Rightarrow{df}=\left({y}−\mathrm{1}\right){dx}_{\mathrm{1}} +{ydx}_{\mathrm{2}} +\left({x}_{\mathrm{1}} +{x}_{\mathrm{2}} \right){dy} \\ $$
Answered by mr W last updated on 01/Jun/24
$${dy}=\frac{{x}_{\mathrm{2}} {dx}_{\mathrm{1}} }{\left({x}_{\mathrm{1}} +{x}_{\mathrm{2}} \right)^{\mathrm{2}} }−\frac{{x}_{\mathrm{1}} {dx}_{\mathrm{2}} }{\left({x}_{\mathrm{1}} +{x}_{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$