Question

Частная производная dz/dy функции z=arctg(x+y/x-y) в точке (-12;5) равна

Answer 1

$\frac{dx}{dy}=arctg(\frac{x+y}{x-y})'*\frac{x+y}{x-y}'=\\\\ \frac{1}{1+(\frac{x+y}{x-y})^2}*\frac{(x+y)'(x-y)-(x+y)(x-y)'}{(x-y)^2}\\\\ \frac{1}{1+\frac{x^2+2xy+y^2}{x^2-2xy+y^2}}*\frac{2x}{(x-y)^2}=\frac{x}{x^2+y^2}\\\\ \frac{-12}{144+25}=\frac{-12}{169}$