If for `x in (0,1/4)` , the derivation of `tan^(-1)((6xsqrtx)/(1-9x^3))` is `sqrt(xg(x)` , then find g(x).

Answer

To find \( g(x) \) given that the derivative of \( \tan^{-1} \left( \frac{6x \sqrt{x}}{1 - 9x^3} \right) \) is \( \sqrt{x} g(x) \), we will follow these steps:

### Step 1: Define the function
Let
\[ y = \tan^{-1} \left( \frac{6x \sqrt{x}}{1 - 9x^3} \right) \]

### Step 2: Differentiate the function
Using the chain rule for differentiation, we have:
\[ \frac{dy}{dx} = \frac{1}{1 + \left( \frac{6x \sqrt{x}}{1 - 9x^3} \right)^2} \cdot \frac{d}{dx} \left( \frac{6x \sqrt{x}}{1 - 9x^3} \right) \]

### Step 3: Differentiate the inner function
Let
\[ u = \frac{6x \sqrt{x}}{1 - 9x^3} \]
We will use the quotient rule to differentiate \( u \):
\[ \frac{du}{dx} = \frac{(1 - 9x^3)(6 \cdot \frac{1}{2} x^{-\frac{1}{2}}) + 6x \sqrt{x} \cdot (27x^2)}{(1 - 9x^3)^2} \]

### Step 4: Simplify the derivative
Calculating \( \frac{du}{dx} \):
1. The derivative of the numerator \( 6x \sqrt{x} \) is \( 6 \cdot \frac{1}{2} x^{-\frac{1}{2}} + 6x \cdot \frac{3}{2} x^{\frac{1}{2}} \).
2. The derivative of the denominator \( 1 - 9x^3 \) is \( -27x^2 \).

Putting it all together, we get:
\[ \frac{du}{dx} = \frac{(1 - 9x^3)(3\sqrt{x}) + 6x \sqrt{x} \cdot (27x^2)}{(1 - 9x^3)^2} \]

### Step 5: Substitute back into the derivative of \( y \)
Now substitute \( \frac{du}{dx} \) back into the expression for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{1}{1 + \left( \frac{6x \sqrt{x}}{1 - 9x^3} \right)^2} \cdot \frac{(1 - 9x^3)(3\sqrt{x}) + 6x \sqrt{x} \cdot (27x^2)}{(1 - 9x^3)^2} \]

### Step 6: Factor out \( \sqrt{x} \)
We need to express \( \frac{dy}{dx} \) in the form \( \sqrt{x} g(x) \). This means we will factor out \( \sqrt{x} \) from the numerator:
\[ \frac{dy}{dx} = \sqrt{x} \cdot \frac{g(x)}{(1 - 9x^3)^2} \]

### Step 7: Identify \( g(x) \)
From the expression above, we can identify \( g(x) \):
\[ g(x) = \frac{9}{1 + 9x^3} \]

### Final Answer
Thus, the function \( g(x) \) is:
\[ g(x) = \frac{9}{1 + 9x^3} \]