Question

Найти производные 2 порядка функции, заданной неявно: y^2-x^2+2y=2

Answer 1

$y^2-x^2+2y=2\\\\2y\cdot y'-2x+2y'=0\\\\2y'(y+1)=2x\\\\y'=\frac{2x}{2(y+1)}=\frac{x}{y+1}\\\\y''= \frac{1\cdot (y+1)-x\cdot y'}{(y+1)^2}= \frac{y+1-x\cdot \frac{x}{y+1}}{(y+1)^2} = \frac{(y+1)^2-x^2}{(y+1)^3} =\frac{(y+1-x)(y+1+x)}{(y+1)^3}$