1x+1
1(x+1)2
1−xx+1
1−x(x+1)2
Correct answer is B
y=xx+1
Using quotient rule because the function is of the form u(x)v(x)
dydx=vdudx−udvdxv2
dydx=(x+1).1−x.1(x+1)2
= 1(x+1)2
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