от Добромир Глухаров » 08 Сеп 2017, 17:40
$z=x.arcsin(y^2-x^2)$
$f(u)=arcsin(u)$
$f'(u)=\frac{u'}{\sqrt{1-u^2}}$
$z'_x=\frac{\partial z}{\partial x}=(x)'_x.arcsin(y^2-x^2)+x\cdot\frac{(y^2-x^2)'_x}{\sqrt{1-(y^2-x^2)^2}}=$
$=1.arcsin(y^2-x^2)+\frac{x.(-2x)}{\sqrt{1-(y^2-x^2)^2}}=arcsin(y^2-x^2)-\frac{2x^2}{\sqrt{1-(y^2-x^2)^2}}$
$z'_y=\frac{\partial z}{\partial y}=(x)'_y.arcsin(y^2-x^2)+x\cdot\frac{(y^2-x^2)'_y}{\sqrt{1-(y^2-x^2)^2}}=$
$=0.arcsin(y^2-x^2)+\frac{x.(2y)}{\sqrt{1-(y^2-x^2)^2}}=\frac{2xy}{\sqrt{1-(y^2-x^2)^2}}$