Question

Найти производные dy/dx функции

Answer 1

Ответ.

 $x=y^2+arctg(xy)\ \ \ \Rightarrow \ \ \ y^2=x-arctg(xy)\\\\\\2yy'=1-\dfrac{1}{1+(xy)^2}\cdot (y+xy')\\\\\\2yy'=1-\dfrac{y}{1+(xy)^2}-\dfrac{xy'}{1+(xy)^2}\\\\\\2yy'+\dfrac{xy'}{1+(xy)^2}=1-\dfrac{y}{1+(xy)^2}\\\\\\y'\cdot \Big(2y+\dfrac{x}{1+(xy)^2}\Big)=1-\dfrac{y}{1+(xy)^2}\\\\\\y'\cdot \dfrac{2y+2x^2y^3+x}{1+(xy)^2}=\dfrac{1+x^2y^2-y}{1+(xy)^2}\\\\\\y'=\dfrac{1+x^2y^2-y}{1+(xy)^2}\cdot \dfrac{1+(xy)^2}{2y+2x^2y^3+x}\\\\\\y'=\dfrac{1+x^2y^2-y}{2y+2x^2y^3+x}$