Question

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

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

Answer 1

The correct option is C

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

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

$\therefore y=\left({\tan }^{-1}\right(x+1)-{\tan }^{-1}x)+\left({\tan }^{-1}\right(x+2)-{\tan }^{-1}(x+1\left)\right)+\dots +\left({\tan }^{-1}\right(x+n)-{\tan }^{-1}n)$

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

$\Rightarrow \frac{dy}{dx}=\frac{1}{1+(x+n{)}^2}-\frac{1}{1+x^2}$

$\therefore {\left(\frac{dy}{dx}\right)}_{x=0}=-\frac{n^2}{1+n^2}$