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rosijanka [135]
3 years ago
12

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

Mathematics
1 answer:
notsponge [240]3 years ago
8 0
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


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