Question

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

• A. $\frac{-1}{1+n^2}$
• B. $\frac{-n^2}{1+n^2}$
• C. $\frac{n^2}{1+n^2}$
• D. $n$

Answer 1

The correct option is B $\frac{-n^2}{1+n^2}$

Consider the given expression.

$y={\tan }^{-1}\left(\frac{1}{1+x+x^2}\right)+{\tan }^{-1}\left(\frac{1}{x^2+3x+3}\right)+....+nterms$

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

We know that,

${\tan }^{-1}A-{\tan }^{-1}B={\tan }^{-1}\left(\frac{A-B}{1+AB}\right)$

Therefore,

$y={\tan }^{-1}(x+1)-{\tan }^{-1}x+{\tan }^{-1}(x+2)-{\tan }^{-1}(x+1)+.....+{\tan }^{-1}(x+n)-{\tan }^{-1}[x+(n-1)]$

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

Differentiate $y$ with respect to $x$.

$y^{\prime }=\frac{1}{1+{(x+n)}^2}-\frac{1}{1+x^2}$

Therefore,

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

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

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

Hence, this is the required result.