Question

Найти дифференциалы первого и второго порядков:

Answer 1

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

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