Question

Найти дифференциалы первого и второго порядков:

Answer 1

$z=x^4+3xy^4-2\sqrt{y}\\\\\frac{\partial z}{\partial x}=4x^3+3y^4\; \; \; ,\; \; \; \; \frac{\partial z}{\partial y}=3x\cdot 4y^3-2\cdot \frac{1}{2\sqrt{y}}=12xy^3-\frac{1}{\sqrt{y}}\\\\\\dz=\frac{\partial z}{\partial x}\cdot dx+\frac{\partial z}{\partial y}=(4x^3+3y^4)\cdot dx+(12xy^3-\frac{1}{\sqrt{y}})\cdot dy\\\\\\\frac{\partial ^2z}{\partial x^2}=12x^2\; ,\; \; \frac{\partial ^2z}{\partial x\, \partial y}=12y^3\; \; ,\; \; \frac{\partial ^2z}{\partial ^2y}=36xy^2-\frac{1}{2\sqrt{y^3}}$

$\frac{\partial ^2z}{\partial x^2}=12x^2\; ,\; \; \frac{\partial ^2z}{\partial x\, \partial y}=12y^3\; \; ,\; \; \frac{\partial ^2z}{\partial ^2y}=36xy^2-\frac{1}{2\sqrt{y^3}}\\\\\\d^2z=\frac{\partial ^2z}{\partial x^2}\cdot dx^2+2\frac{\partial ^2z}{\partial x\partial y}\cdot dx\, dy+\frac{\partial ^2z}{\partial y^2}\cdot dy^2\\\\\\d^2z=12x^2\cdot dx^2+12y^3\cdot dx\, dy+(36xy^2-\frac{1}{2\sqrt{y^3}})\cdot dy^2$