If
\(y=\tan ^{-1}\left(\frac{1}{1+x+x^{2}}\right)+\tan ^{-1}\left(\frac{1}{x^{2}+2 x+3}\right)\)
\(+\tan ^{-1}\left(\frac{1}{x^{2}+5 x+7}\right)+\ldots+n\) terms, then \(y^{\prime}(0)\) is

Given,
\[
\begin{aligned}
y &=\tan ^{-1}\left(\frac{1}{x^{2}+x+1}\right)+\tan ^{-1}\left(\frac{1}{x^{2}+2 x+3}\right) \\
&+\tan ^{-1}\left(\frac{1}{x^{2}+5 x+7}\right)+\ldots n \text { terms } \ldots(
\end{aligned}
\]
\(\operatorname{Now}, \tan ^{-1}\left(\frac{1}{x^{2}+x+1}\right)=\left(\frac{x+1-x}{1+x(x+1)}\right)\)
\[
=\tan ^{-1}(\mathrm{x}+1)-\tan ^{-1} \mathrm{x}
\]
From Eq. (i), we get
\[
\begin{aligned}
&y=\left[\tan ^{-1}(x+1)-\tan ^{-1} x\right]+\left[\tan ^{-1}(x+2)\right. \\
&\left.-\tan ^{-1}(x+1)\right]+\ldots+\left[\tan ^{-1}(x+n)-\tan ^{-1} \mathrm{n}\right] \\
&\therefore \quad y=\tan ^{-1}(x+n)-\tan ^{-1}(x)
\end{aligned}
\]
On differentiating w.r.t. ' \(x\) ', we get
\[
\begin{aligned}
\therefore \quad \frac{\mathrm{dy}}{\mathrm{dx}} &=\frac{1}{1+(\mathrm{x}+\mathrm{n})^{2}}-\frac{1}{1+\mathrm{x}^{2}} \\
\therefore \mathrm{dx})_{\text {at } \mathrm{x}=0} &=\frac{1}{1+\mathrm{n}^{2}}-1 \\
&=\frac{1-\left(1+\mathrm{n}^{2}\right)}{1+\mathrm{n}^{2}} \\
&=\frac{-\mathrm{n}^{2}}{1+\mathrm{n}^{2}}
\end{aligned}
\]