If y = tan^{-1}(frac{1}{x^2+x+1}) + tan^{-1}( | vedclass

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(B) The $k$-th term of the series is $T_k = 	an^{-1}(\frac{1}{x^2 + (2k-1)x + (k^2-k+1)})$.We can write the denominator as $1 + (x+k-1)(x+k)$.Thus, $

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$-\frac{n^2}{n^2+1}$

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(B) The $k$-th term of the series is $T_k = 	an^{-1}(\frac{1}{x^2 + (2k-1)x + (k^2-k+1)})$.We can write the denominator as $1 + (x+k-1)(x+k)$.Thus, $T_k = 	an^{-1}(\frac{(x+k) - (x+k-1)}{1 + (x+k)(x+k-1)}) = 	an^{-1}(x+k) - 	an^{-1}(x+k-1)$.The sum $y$ is a telescoping series: $y = \sum_{k=1}^{n} (	an^{-1}(x+k) - 	an^{-1}(x+k-1)) = 	an^{-1}(x+n) - 	an^{-1}(x)$.Differentiating with respect to $x$: $y' = \frac{1}{1+(x+n)^2} - \frac{1}{1+x^2}$.At $x=0$, $y'(0) = \frac{1}{1+n^2} - \frac{1}{1+0^2} = \frac{1}{1+n^2} - 1 = \frac{1 - (1+n^2)}{1+n^2} = -\frac{n^2}{n^2+1}$.

List-$I$	List-$II$
$A$. $\sec^{-1} x$	$I$. $\frac{1}{1-x^2}, x \in (-1, 1)$
$B$. $\tanh^{-1} x$	$II$. $\frac{-1}{\|x\| \sqrt{x^2+1}}, x \neq 0$
$C$. $\coth^{-1} x$	$III$. $\frac{1}{\|x\| \sqrt{x^2-1}}, \|x\| > 1$
$D$. $\operatorname{cosech}^{-1} x$	$IV$. $\frac{1}{1-x^2}, x \in R - [-1, 1]$
	$V$. $\frac{-1}{\|x\| \sqrt{1-x^2}}, \|x\| < 1, x \neq 0$

If $y = \tan^{-1}(\frac{1}{x^2+x+1}) + \tan^{-1}(\frac{1}{x^2+3x+3}) + \tan^{-1}(\frac{1}{x^2+5x+7}) + \dots$ upto $n$ terms, then $y'(0) = $

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