Log(2)5*1/log(2)5+1/6log(3)3=1+1/6=1 1/6
См фото
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Как-то так вроде.
f'(y)=(((x-1)/(x+1))^2)' * ((x-1)/(x+1))'=2((x-1)/(x+1)) * ((x-1)'(x+1)-(x-1)(x+1)')/(x+1)^2=2((x-1)/(x+1)) * ((1-0)(x+1)-(x-1)(1+0))/(x+1)^2=2((x-1)/(x+1)) * (x+1-x+1)/(x+1)^2=2((x-1)/(x+1)) *2/(x+1)^2