Animated Solution for Mathematics - Differentiation: 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

Welcome, fellow traveler of the mathematical landscape. Today, we stand before a function that, at first glance, seems designed to intimidate.

We see the function:

It is a nesting doll of operations—a sine, an inverse cosine, and an exponential fraction. It is easy to feel overwhelmed, but I want you to take a deep breath. In the world of JEE Advanced, complexity is often just a mask. Our job is to peel it away, layer by layer, to reveal the simple, elegant truth underneath.

Look closely at the inner term: $\frac{1-2^{2x}}{1+2^{2x}}$. Does it stir a memory? It should.

It bears a striking resemblance to the trigonometric identity:

This is our golden key. By setting $2^x=\tan \theta$, we are not just simplifying; we are translating the problem from the language of algebra into the language of geometry.

Since $2^x$ is always positive, our angle $\theta$ is safely tucked away in the interval $(0,\pi /2)$. This geometric model is robust, and it ensures that our subsequent steps are mathematically sound.

With our substitution $2^x=\tan \theta$, the expression $\frac{1-2^{2x}}{1+2^{2x}}$ transforms into $\cos 2\theta$. Now, our function becomes:

Here is where the magic happens. Because $\theta \in (0,\pi /2)$, we know that $2\theta \in (0,\pi )$.

This is the principal range of the inverse cosine function! The inverse cosine and cosine neutralize each other, leaving us with the beautifully simple $f(x)=\sin 2\theta$. The mountain of complexity has vanished, leaving us with a clear path forward.

We are not done yet. We must return to the world of $x$. Using the identity $\sin 2\theta =\frac{2\tan \theta }{1+{\tan }^2\theta }$ and substituting back $\tan \theta =2^x$, we get:

Now, we apply the quotient rule. I know the quotient rule can be tedious, but treat it like a dance—step by step, rhythmically.

We differentiate the numerator and denominator, keeping that crucial $\ln 2$ factor in mind for the exponential terms. When we evaluate the derivative at $x=1$, we find:

We compare our result with the given form $-\frac{b}{a}\ln 2$. It is clear that $b=12$ and $a=25$.

The final task is to calculate $|a^2-b^2|$, which is:

We have arrived at the destination. This problem was never about brute force; it was about recognizing patterns, respecting domains, and trusting the elegance of mathematics. You have conquered the complexity, and the final answer is 481.