Question

Буду благодарен если поможете

Answer 1

Ответ:

$z=arctg\dfrac{x+y}{1+xy}\ \ ,\ \ \ x=t^2+1\ \ ,\ \ y=sint\\\\\\\boxed {\ z=z\Big(x(t)\ ;\ y(t)\Big)\ \ ,\ \ \ \ \dfrac{dz}{dt}=\dfrac{\partial z(x;y)}{\partial x}\cdot \dfrac{dx}{dt}+\dfrac{\partial z(x;y)}{\partial y}\cdot \dfrac{dy}{dt}\ }$

$\dfrac{dz}{dt}=\dfrac{1}{1+\dfrac{(x+y)^2}{(1+xy)^2}}\cdot \dfrac{1+xy-y(x+y)}{(1+xy)^2}\cdot 2t+\\\\\\+\dfrac{1}{1+\dfrac{(x+y)^2}{(1+xy)^2}}\cdot \dfrac{1+xy-x(x+y)}{(1+xy)^2}\cdot cost=\\\\\\=\dfrac{1-y^2}{(1+xy)^2+(x+y)^2}\cdot 2t+\dfrac{1-x^2}{(1+xy)^2+(x+y)^2}\cdot cost$