Question

Найти диференциал $Y = \frac{dY}{dX_1} + \frac{dY}{dX_2} + \frac{dY}{dX_3}$ от функции $Y= X_3 + \frac{X_1 + X_3}{\sqrt{X_1}} }$

Answer 1

Ответ:

$\displaystyle Y=x_3+\frac{x_1+x_3}{\sqrt{x_1}}=x_3+\sqrt{x_1}+\frac{1}{\sqrt{x_1}}\cdot x_3\\\\\\\boxed{\ dY=\frac{\partial Y}{\partiai x_1}\cdot dx_1+\frac{\partial Y}{\partiai x_2}\xsot dx_2+\frac{\partial Y}{\partiai x_3}\cdot dx_3\ }\\\\\\\frac{\partial Y}{\partiai x_1}=0+\frac{1}{2\sqrt{x_1}}-\frac{1}{2}\cdot x_1^{-\frac{3}{2}}\cdot x_3=\frac{1}{2\sqrt{x_1}}-\frac{x_3}{2\sqrt{x_1^3}}\\\\\\\frac{\partial Y}{\partiai x_2}=0$

$\displaystyle \frac{\partial Y}{\partiai x_3}=1+0+\frac{1}{\sqrt{x_1}}=1+\frac{1}{\sqrt{x_1}}\\\\\\dY=\Big(\frac{1}{2\sqrt{x_1}}-\frac{x_3}{2\, \sqrt{x_1^3}}\Big)\, dx_1+\Big(1+\frac{1}{\sqrt{x_1}}\Big)\, dx_3$