Question

If $y=\sum _{r=1}^x{\tan }^{-1}\frac{1}{1+r+r^2}$ and $\frac{dy}{dx}$ is equal to

• A. $\frac{1}{1+x^2}$
• B. $\frac{1}{1+(1+x{)}^2}$
• C. $0$
• D. None of these

Answer 1

Answer: (B)

$\frac{1}{1+(1+x{)}^2}$

$y=\sum _{r=1}^x{\tan }^{-1}\frac{1}{1+r+r^2}$

$={\sum }_{r=1}^x{\tan }^{-1}\frac{(r+1)-r}{1+r(r+1)}$
$=\sum _{r=1}^x\left[{\tan }^{-1}\right(r+1)-{\tan }^{-1}r]$

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