If   $$\begin{gathered} y={\tan }^{-1}\frac{1}{1+x+x^2}+{\tan }^{-1}\frac{1}{x^2+3x+3} \\ +{\tan }^{-1}\frac{1}{x^2+5x+7}+{\tan }^{-1}\frac{1}{x^2+7x+13}+\mathrm{...}+n terms \end{gathered}$$

 then $y^{\text{'}}\left(0\right)$  is equal to 

$y={\tan }^{-1}\frac{1}{1+x+x^2}+{\tan }^{-1}\frac{1}{x^2+3x+3}+\mathrm{...}+$upto   terms

$={\tan }^{-1}\frac{\left(x+1\right)-x}{1+x\left(x+1\right)}+{\tan }^{-1}\frac{\left(x+2\right)-\left(x+1\right)}{1+\left(x+1\right)\left(x+2\right)}+\mathrm{...}n$terms 

$={\tan }^{-1}\left(x+1\right)-{\tan }^{-1}x+{\tan }^{-1}\left(x+2\right)-{\tan }^{-1}\left(x+1\right)+\mathrm{...}+$

${\tan }^{-1}\left(x+n\right)-{\tan }^{-1}\left(x+\left(n-1\right)\right)$

$={\tan }^{-1}\left(x+n\right)-{\tan }^{-1}x.$

$y^{\text{'}}\left(x\right)=\frac{1}{1+{\left(x+n\right)}^2}-\frac{1}{1+x^2}$

$\Rightarrow y^{\text{'}}\left(0\right)=\frac{1}{1+n^2}-1=-\frac{n^2}{1+n^2}$