simplify: $\displaystyle \tan\left(2\cos^{-1}(\frac{x}{4})\right)$

using the double angle foruma $\displaystyle tan(2x) = \frac{2tan(x)}{1-tan^2x}$

i got to this part:

$\displaystyle \frac{2tan(\cos^{-1}(\frac{1}{4x}))} {1-tan(\cos^{-1}(\frac{1}{4x}))^2}$