MuhammadAzhanFikry MuhammadAzhanFikry
  • 12-10-2016
  • Mathematics
contestada

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

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apologiabiology
apologiabiology apologiabiology
  • 12-10-2016
f'[tex]( \frac{y-1}{y+1})-y [/tex]=
f'[tex]( \frac{y-1}{y+1}) [/tex]-f'(y) [/tex]=
[tex]\frac{f'(y-1)(y+1)-f'(y+1)(y-1)}{(y+1)^2} -1[/tex]=
[tex]\frac{(f'(y)-f'(1))(y+1)-(f'(y)+f'(1))(y-1)}{(y+1)^2} -1[/tex]=
[tex]\frac{(1-0)(y+1)-(1-0)(y-1)}{(y+1)^2} -1[/tex]=
[tex]\frac{(1)(y+1)-(1)(y-1)}{(y+1)^2} -1[/tex]=
[tex]\frac{(y+1)-(y-1)}{(y+1)^2} -1[/tex]=
[tex]\frac{y+1-y+1}{(y+1)^2} -1[/tex]=
[tex]\frac{2}{(y+1)^2} -1[/tex]


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