If $$f(x) = \sin\left(\cos^{-1}\left(\frac{1-2^{2x}}{1+2^{2x}}\right)\right)$$ and its first derivative with respect to $$x$$ is $$-\frac{b}{a}\log_e 2$$ when $$x = 1$$, where $$a$$ and $$b$$ are integers, then the minimum value of $$|a^2 - b^2|$$ is ________.

We have $$f(x) = \sin\!\left(\cos^{-1}\!\left(\frac{1-2^{2x}}{1+2^{2x}}\right)\right)$$.

Let $$t = 2^x$$, so $$2^{2x} = t^2$$. The argument becomes $$\frac{1-t^2}{1+t^2}$$.

Using the identity $$\cos 2\theta = \frac{1-\tan^2\theta}{1+\tan^2\theta}$$, if we set $$\tan\theta = t$$ (with $$t > 0$$, so $$\theta \in (0, \pi/2)$$), then $$\cos^{-1}\!\left(\frac{1-t^2}{1+t^2}\right) = 2\theta$$.

Therefore $$f(x) = \sin(2\theta) = \frac{2\tan\theta}{1+\tan^2\theta} = \frac{2t}{1+t^2} = \frac{2 \cdot 2^x}{1+2^{2x}}$$.

Differentiating with $$u = 2^x$$: $$f = \frac{2u}{1+u^2}$$, so $$\frac{df}{du} = \frac{2(1+u^2) - 2u(2u)}{(1+u^2)^2} = \frac{2-2u^2}{(1+u^2)^2}$$.

By the chain rule, $$f'(x) = \frac{2(1-u^2)}{(1+u^2)^2} \cdot u\ln 2$$, since $$\frac{du}{dx} = 2^x \ln 2 = u\ln 2$$.

At $$x = 1$$: $$u = 2$$, so $$f'(1) = \frac{2(1-4)}{(1+4)^2} \cdot 2\ln 2 = \frac{-6}{25} \cdot 2\ln 2 = -\frac{12}{25}\ln 2$$.

Comparing with $$f'(1) = -\frac{b}{a}\log_e 2$$: we get $$\frac{b}{a} = \frac{12}{25}$$. Since $$\gcd(12,25) = 1$$, the integers with minimum $$|a^2-b^2|$$ are $$a = 25, b = 12$$.

$$|a^2 - b^2| = |625 - 144| = 481$$.

The answer is 481.