Question

u=lntg(y/x) найти d2u/dx2

Answer 1

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