Question

If $y=\sum _{r=1}^x{\tan }^{-1}\frac{1}{1+r+r^2}$ then $\frac{dy}{dx}$ 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)

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

We have,

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

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

$=[{\tan }^{-1}2-{\tan }^{-1}1+{\tan }^{-1}3-{\tan }^{-1}2+...+{\tan }^{-1}x-{\tan }^{-1}(x-1)+{\tan }^{-1}(x+1)-{\tan }^{-1}x]$

$=[{\tan }^{-1}(x+1)-{\tan }^{-1}1]$

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