Question

If $f\left(x\right)=\sum _{k=1}^x{\tan }^{-1}\left(\frac{2k}{2+k^2+k^4}\right),$ then the value of $6f^{\prime }\left(0\right)$ is

Answer 1

$f\left(x\right)={\tan }^{-1}\left(\frac{2k}{2+k^2+k^4}\right)$
$={\tan }^{-1}\left(\frac{2k}{1+(k^4+1+2k^2-k^2)}\right)$
$={\tan }^{-1}\left(\frac{2k}{1+{(k^2+1)}^2-k^2}\right)$
$={\tan }^{-1}\left(\frac{(k^2+1+k)-(k^2+1-k)}{1+(k^2+1+k)(k^2+1-k)}\right)$

$\therefore f\left(x\right)=\sum _{k=1}^x{\tan }^{-1}(k^2+1+k)-{\tan }^{-1}(k^2+1-k)$

Putting the values of $k,$ we get
$f\left(x\right)={\tan }^{-1}\left(3\right)-{\tan }^{-1}\left(1\right)+{\tan }^{-1}\left(7\right)-{\tan }^{-1}\left(3\right)+\cdots +{\tan }^{-1}(x^2+x+1)-{\tan }^{-1}(x^2-x+1)$
$\Rightarrow f\left(x\right)={\tan }^{-1}(x^2+x+1)-{\tan }^{-1}\left(1\right)$
Differentiating both sides w.r.t. $x,$
$f^{\prime }\left(x\right)=\frac{1}{1+(x^2+x+1{)}^2}\times (2x+1)$
$\therefore f^{\prime }\left(0\right)=\frac{1}{2}$
Hence, $6f^{\prime }\left(0\right)=3$