Question

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

• A. $-\frac{x^2}{1+x^2}$
• B. $-\frac{n^2}{1+n^2}$
• C. $1$
• D. $0$

Answer 1

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

$y={\text{tan}}^{-1}(\frac{1}{1+x+x^2})+{\text{tan}}^{-1}(\frac{1}{x^2+3x+3})+\cdots n$ terms

$y={\text{tan}}^{-1}(\frac{x+1-x}{1+x(x+1)})+{\text{tan}}^{-1}(\frac{x+2-x-1}{1+(x+1)(x+2)})+\cdots n$ terms

$y={\text{tan}}^{-1}(x+1)-{\text{tan}}^{-1}x+{\text{tan}}^{-1}(x+2)-{\text{tan}}^{-1}(x+1)+\cdots +{\text{tan}}^{-1}(x+n)-{\text{tan}}^{-1}(x+n-1)$

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

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

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