Question

If $y=\sum _{r=1}^x{\tan }^{-1}\left(\frac{1}{1+r+r^2}\right),$ then $\frac{dy}{dx}$ at $x=2$ is equal to

• A. $\frac{1}{5}$
• B. $\frac{1}{10}$
• C. $0$
• D. $\frac{1}{2}$

Answer 1

The correct option is B $\frac{1}{10}$

$y=\sum _{r=1}^x{\tan }^{-1}\frac{1}{1+r+r^2}=\sum _{r=1}^x{\tan }^{-1}\frac{r+1-r}{1+r(1+r)}=\sum _{r=1}^x\left[{\tan }^{-1}\right(r+1)-{\tan }^{-1}(r\left)\right]={\tan }^{-1}(x+1)-{\tan }^{-1}\left(1\right)\Rightarrow \frac{dy}{dx}=\frac{1}{1+(1+x{)}^2}$
$y^{\prime }\left(2\right)=\frac{1}{10}$