Question

Помогите с решением, высшая математика ​

Answer 1

Решение.

Производные сложных функций нескольких переменных.

$\displaystyle \bf z=arctg\frac{u}{v}\ \ ,\ \ u=xy\ ,\ v=1+x^2+y^2\\\\\\z=z(\, u(x,y)\ ;\ v(x,y)\, )\ \ ,\\\\\frac{\partial z}{\partial x}=\frac{\partial z}{\partial u}\cdot \frac{\partial u}{\partial x}+\frac{\partial z}{\partial v}\cdot \frac{\partial v}{\partial x}\ \ ,\ \ \ \frac{\partial z}{\partial y}=\frac{\partial z}{\partial u}\cdot \frac{\partial u}{\partial y}+\frac{\partial z}{\partial v}\cdot \frac{\partial v}{\partial y}$  

$\displaystyle \bf \frac{\partial z}{\partial u}=\frac{1}{1+\frac{u^2}{v^2}} \cdot \frac{1}{v}=\frac{v}{u^2+v^2}\ \ ,\ \ \ \ \frac{\partial u}{\partial x}=y\ \ ,\ \ \ \ \frac{\partial u}{\partial y}=x\\\\\\\frac{\partial z}{\partial v}=\frac{1}{1+\frac{u^2}{v^2}}\cdot \frac{-u}{v^2}=-\frac{u}{u^2+v^2}\ \ ,\ \ \ \ \frac{\partial v}{\partial x}=2x\ \ ,\ \ \ \ \frac{\partial v}{\partial x}=2y\\\\\\\frac{\partial z}{\partial x}=\frac{v}{u^2+v^2}\cdot y-\frac{u}{u^2+v^2}\cdot 2x=\frac{1}{u^2+v^2}\cdot (vy-2ux)$

$\bf \displaystyle \frac{\partial z}{\partial y}=\frac{v}{u^2+v^2}\cdot x-\frac{u}{u^2+v^2}\cdot 2y=\frac{1}{u^2+v^2}\cdot (vx-2uy)$