Предмет: Математика, автор: maxtehas07

Срочно пожалуйста!!!

Знайдіть частинні похідні функції

z(x, y) =sin(x^2+y^2)

Ответы

Автор ответа: polarkat

$\cfrac{\partial }{\partial x}\sin \left ( x^2+y^2 \right )=\cos \left ( x^2+y^2 \right )\cfrac{\partial }{\partial x}\left ( x^2+y^2 \right )=2x\cos \left ( x^2+y^2 \right )\\$

$\cfrac{\partial }{\partial y}\sin \left ( x^2+y^2 \right )=\cos\left ( x^2+y^2 \right )\cfrac{\partial }{\partial x}\left ( x^2+y^2 \right )=2y\cos \left ( x^2+y^2 \right )\\$

$\cfrac{\partial^2 }{\partial x^2}\sin \left ( x^2+y^2 \right )=\cfrac{\partial }{\partial x}\left ( 2x\cos \left ( x^2+y^2 \right ) \right )=2\left (\cos\left ( x^2+y^2 \right )\cfrac{\partial }{\partial x}x+x\cfrac{\partial }{\partial x}\left ( \cos\left ( x^2+y^2 \right ) \right ) \right )=\\=2\left ( \cos\left ( x^2+y^2 \right )-2x^2\sin \left ( x^2+y^2 \right ) \right )\\$

$\cfrac{\partial^2 }{\partial y^2}\sin \left ( x^2+y^2 \right )=\cfrac{\partial }{\partial y}2y\cos\left ( x^2+y^2 \right )=2\left ( \cos\left ( x^2+y^2 \right )\cfrac{\partial }{\partial y}y+y\cfrac{\partial }{\partial y}\cos \left ( x^2+y^2 \right ) \right )=\\=2\left ( \cos\left ( x^2+y^2 \right )-2y^2\sin\left ( x^2+y^2 \right ) \right )$

$\cfrac{\partial ^2}{\partial x\partial y}\sin\left ( x^2+y^2 \right )=\cfrac{\partial }{\partial y}\left ( \cfrac{\partial }{\partial x} \sin\left ( x^2+y^2 \right )\right )=\cfrac{\partial }{\partial y}\left ( 2x\cos\left ( x^2+y^2 \right ) \right )=\\=2x\cfrac{\partial }{\partial y}\cos\left ( x^2+y^2 \right )=-2x\sin\left ( x^2+y^2 \right )\cfrac{\partial }{\partial y}\left ( x^2+y^2 \right )=-4xy\sin\left ( x^2+y^2 \right )$