Chapter 7 - Section 7.3 - Double-Angle, Half-Angle, and Product-Sum Formulas - 7.3 Exercises - Page 561: 44

$\frac{2x\sqrt {1-x^2}}{2x^2-1}$

Let $cos^{-1}x=y$, we have $cos(y)=x, sin(y)=\sqrt {1-x^2}, tan(y)=\frac{\sqrt {1-x^2}}{x}$ Thus $tan(2cos^{-1}x)=tan(2y)=\frac{2tan(y)}{1-tan^2y}=\frac{2\sqrt {1-x^2}}{x}/(1-\frac{1-x^2}{x^2}) =\frac{2x\sqrt {1-x^2}}{2x^2-1}$