Question

Найти dz/dx и dz/dy для y=cos^2x/2y

Answer 1

$z=cos^2\frac{x}{2y}\\\\z'_{x}=2\, cos\frac{x}{2y}\cdot sin\frac{x}{2y}\cdot \frac{1}{2y}=\frac{1}{2y}\cdot sin\frac{x}{y} \\\\z'_{y}=2\, cos\frac{x}{2y}\cdot sin\frac{x}{2y}\cdot \frac{x}{2}\cdot (-\frac{1}{y^2}) =-\frac{x}{2y^2}\cdot sin\frac{x}{y}$