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 , $\frac{dy}{dx}_{x=0} = - \frac{1}{2}$