Differentiation

For such type of questions try to find tr

t_{r} = tan^{-1}\frac{1}{1+(x+r-1)(x+r)}

t_{r} = tan^{-1}\frac{(x+r) - (x+r-1)}{1+(x+r-1)(x+r)}

t_{r} = tan^{-1}\(x+r) - tan^{-1}(x+r-1)

summing this from r=1 to r=n

y = tan^{-1}\(x+n) - tan^{-1}(x)

y = tan^{-1}\frac{n}{1+(x+n)x}

Now can you proceed further?

yeah,so now on differentiating we get y'(x)=(1/1+(x+n)²) - (1/1+x²).thus y'(0)=-n²/(1+n²) ,that is option b.