Animated Solution for Mathematics - Differentiation: If for $x\in (0,\frac{1}{4})$, the derivative of ${\tan }^{-1}\left(\frac{6x\sqrt{x}}{1-9x^3}\right)$ is $\sqrt{x}\cdot g(x)$, then $g(x)$ equals:

Imagine you are standing before a complex, intimidating expression: ${\tan }^{-1}\left(\frac{6x\sqrt{x}}{1-9x^3}\right)$. At first glance, it looks like a trap—a labyrinth of radicals and powers designed to make you stumble.

But in the world of JEE Advanced, intimidation is often a mask for elegance. Our mission is to peel back that mask.

The first step in our journey is to stop looking at the expression as a whole and start looking at its soul. We see $6x\sqrt{x}$ in the numerator and $1-9x^3$ in the denominator.

If you have spent enough time with trigonometry, your brain should immediately start ringing bells. Does this not look like the double-angle identity for tangent?

Recall that:

If we can force our expression into this form, the inverse tangent will simply vanish, leaving us with a much friendlier $2\theta$.

Let us perform a surgical strike on the expression. We notice that $9x^3$ is the square of $3x^{3/2}$. Similarly, $6x\sqrt{x}$ is $2\cdot 3x^{3/2}$.

This is the 'Aha!' moment. By defining a new variable $u=3x^{3/2}$, our complex fraction transforms into the beautiful, simple form:

Now, the function becomes $y={\tan }^{-1}\left(\frac{2u}{1-u^2}\right)$. Because we checked our domain constraints—ensuring $u$ stays within the safe bounds where the identity holds—we can confidently simplify this to $y=2{\tan }^{-1}(u)$.

We have traded a terrifying algebraic fraction for a simple inverse tangent function. This is the essence of mathematical mastery: simplification before calculation.

Now that we have $y=2{\tan }^{-1}(3x^{3/2})$, the differentiation is no longer a chore; it is a victory lap. We apply the chain rule:

The derivative of $3x^{3/2}$ is a straightforward power rule application:

When we multiply this by our previous term, the $2$ in the numerator and the $2$ in the denominator cancel out with poetic precision, leaving us with:

Comparing this to the target form $\sqrt{x}\cdot g(x)$, we can see clearly that $g(x)=\frac{9}{1+9x^3}$. You have successfully navigated the trap, simplified the chaos, and arrived at the truth.