Question

Решите пожалуйста 4ый пример

Answer 1

Ответ:

$z=e^{xy}\\\\\dfrac{\partial z}{\partial x}=e^{xy}\cdot y\ \ ,\ \ \dfrac{\partial z}{\partial y}=e^{xy}\cdot x\ \ ,\\\\\\\dfrac{\partial ^2z}{\partial x^2}=y^2\cdot e^{xy}\ \ ,\ \ \dfrac{\partial ^2z}{\partial x\partial y}=e^{xy}+xy\cdot e^{xy}\ \ ,\ \ \dfrac{\partial ^2z}{\partial y^2}=x^2\cdot e^{xy}\\\\\\\\x^2\cdot \dfrac{\partial ^2z}{\partial x^2}-2xy\cdot \dfrac{\partial ^2z}{\partial x\partial y}+y^2\cdot \dfrac{\partial ^2z}{\partial y^2}+2xyz=$

$=x^2y^2\, e^{xy}-2xy\, e^{xy}\, (1+xy)+x^2y^2\, e^{xy}+2xy\, e^{xy}=\\\\\\=e^{xy}\cdot (2x^2y^2-2xy-2x^2y^2+2xy)=e^{xy}\cdot 0=0$