Автор - KheKhr

Дана функция f(x;y)=sin(x^2y) . Найти все её частные производные второго порядка.

Ответ

Автор - moboqe

Ответ:

frac{partial ^2 f}{partial x^2}=frac{partial}{partial x}left(2xycos{(x^2y)}right)=2ycos{(x^2y)}-2xysin{(x^2y)}cdot2xy=2yleft(cos{(x^2y)}-2x^2ysin{(x^2y)}right)\\\ frac{partial ^2 f}{partial x partial y}=frac{partial}{partial x}left(x^2cos{(x^2y)}right)=-x^2sin{(x^2y)}cdot2xy+2xcos{(x^2y)}=2xleft(cos{(x^2y)}-x^2ysin{(x^2y)}right)\\\ frac{partial ^2 f}{partial y^2}=frac{partial}{partial y}left(x^2cos{(x^2y)}right)=-x^2sin{(x^2y)}cdot x^2=-x^4sin{(x^2y)}\\\ frac{partial ^2 f}{partial y partial x}=frac{partial}{partial y}left(2xycos{(x^2y)}right)=2xcos{(x^2y)}-x^2sin{(x^2y)}cdot2xy=2xleft(cos{(x^2y)}-x^2ysin{(x^2y)}right)\\\

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