Question

$y={\tan }^{-1}\left(\frac{1}{1+x+x^2}\right)+{\tan }^{-1}\left(\frac{1}{x^2+3x+3}\right)+{\tan }^{-1}\left(\frac{1}{x^2+5x+7}\right)+......+$ upto $n$ terms, then $\frac{dy}{dx}$ at $x=0$ and $n=1$ is equal to

• A. $\frac{1}{2}$
• B. $-\frac{1}{2}$
• C. 0
• D. $\frac{1}{3}$

Answer 1

Answer: (B)

For $n=1,y={\tan }^{-1}\left(\frac{1}{1+x+x^2}\right)$
$\therefore \frac{dy}{dx}=\frac{1}{1+{\left(\frac{1}{1+x+x^2}\right)}^2}.\frac{-1}{{\left(1+x+x^2\right)}^2}.\left(2x+1\right)$
$=\frac{\left(-\left(2x+1\right)\right)}{{\left(1+x+x^2\right)}^2+1}$
$\frac{dy}{dx}=-\frac{2x+1}{x^4+2x^3+3x^2+2x+2}$
Now , ${\frac{dy}{dx}}_{x=0}=-\frac{1}{2}$