The correct option is B 110ln 2
We know that ddx(tan−1x)=11+x2 , but in this case the argument of tan−1x is a different function of x and not x itself. So we will apply chain rule.
dydx=11+[2x1+2x+1]2×ddx(2x1+2x+1) [Using chain rule]
=11+[2x1+2x+1]2×2x ln2(1+2x+1)−2x×2x+1ln2(1+2x+1)2 [Using quotient rule]
Put x=0 to get dydx∣∣x=0=110ln2 which is the correct answer