Question

X (dy/dx) = (y-1/y+1)-y

Answer 1

F'$( \frac{y-1}{y+1})-y$=
f'$( \frac{y-1}{y+1})$-f'(y) [/tex]=
$\frac{f'(y-1)(y+1)-f'(y+1)(y-1)}{(y+1)^2} -1$=
$\frac{(f'(y)-f'(1))(y+1)-(f'(y)+f'(1))(y-1)}{(y+1)^2} -1$=
$\frac{(1-0)(y+1)-(1-0)(y-1)}{(y+1)^2} -1$=
$\frac{(1)(y+1)-(1)(y-1)}{(y+1)^2} -1$=
$\frac{(y+1)-(y-1)}{(y+1)^2} -1$=
$\frac{y+1-y+1}{(y+1)^2} -1$=
$\frac{2}{(y+1)^2} -1$