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

User Wix
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1 Answer

3 votes
f'
( (y-1)/(y+1))-y=
f'
( (y-1)/(y+1))-f'(y) [/tex]=

(f'(y-1)(y+1)-f'(y+1)(y-1))/((y+1)^2) -1=

((f'(y)-f'(1))(y+1)-(f'(y)+f'(1))(y-1))/((y+1)^2) -1=

((1-0)(y+1)-(1-0)(y-1))/((y+1)^2) -1=

((1)(y+1)-(1)(y-1))/((y+1)^2) -1=

((y+1)-(y-1))/((y+1)^2) -1=

(y+1-y+1)/((y+1)^2) -1=

(2)/((y+1)^2) -1


User Muhammad Talha
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