For $x<\frac{1}{2}$, derivative of $\tan ^{-1}\left(\frac{1+2 x}{1-2 x}\right)$ with respect to $\sqrt{1+4 x^2}$ is:

$\frac{1}{2 x \sqrt{1+4 x^2}}$

The correct answer is Option (2) → $\frac{1}{2 x \sqrt{1+4 x^2}}$

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

$z=\sqrt{1+4x^2}$

Now,

$\frac{d\tan^{-1}u}{dx}=\frac{1}{1+u^2}.\frac{du}{dx}$, where $u=\frac{1+2x}{1-2x}$

$⇒\frac{du}{dx}=\frac{(1-2x)(2)-(1+2x)(-2)}{(1-2x)^2}$

$=\frac{4}{(1-2x)^2}$

Now, compute

$1+u^2=1+\left(\frac{1+2x}{1-2x}\right)^2$

$=\frac{(1-2x)^2+(1+2x)^2}{(1-2x)^2}$

$=\frac{2(1+4x^2)}{(1-2x)^2}$

$∴\frac{dy}{dx}=\frac{1}{1+u^2}\frac{du}{dx}$

$=\frac{2}{1+4x^2}$

and,

$z=\sqrt{1+4x^2}$

$\frac{dz}{dx}=-\frac{1}{2\sqrt{1+4x^2}}.8x$

$=\frac{4x}{\sqrt{1+4x^2}}$

$∴\frac{dy}{dx}=\frac{dy}{dx}×\frac{dz}{dz}=\frac{1}{2x\sqrt{1+4x^2}}$