1. The problem is to find the derivative of the function $f(x) = \tan^{-1}(2x)$.\n\n2. Recall the formula for the derivative of the inverse tangent function: $$\frac{d}{dx} \tan^{-1}(u) = \frac{1}{1+u^2} \cdot \frac{du}{dx}.$$\n\n3. Here, $u = 2x$, so we need to find $\frac{du}{dx}$. Since $u = 2x$, then $$\frac{du}{dx} = 2.$$\n\n4. Substitute $u$ and $\frac{du}{dx}$ into the derivative formula: $$\frac{d}{dx} \tan^{-1}(2x) = \frac{1}{1+(2x)^2} \cdot 2 = \frac{2}{1+4x^2}.$$\n\n5. Therefore, the derivative of $\tan^{-1}(2x)$ is $$\boxed{\frac{2}{1+4x^2}}.$$