Question

u=lntg(y/x) найти d2u/dx2

Answer 1

$\frac{du}{dx} =-\frac{y}{x^2cos^2(\frac{y}{x})tg(\frac{y}{x})}=-\frac{2y}{x^2sin(\frac{2y}{x})} \\\\ \frac{d^2y}{dx^2}=-2y(\frac{x^{-2}}{sin(\frac{2y}{x})})'=-2y\frac{-2x^{-3}sin(\frac{2y}{x})+cos(\frac{2y}{x})}{sin^2(\frac{2y}{x})}$