Differentiate w.r.t. as indicated :
`"tan"^(-1)(3x-x^(3))/(1-3x^(2))" w.r.t. ""tan"^(-1)(2x)/(1-x^(2))`

Answer

To differentiate the function \( \frac{\tan^{-1}(3x - x^3)}{1 - 3x^2} \) with respect to \( \frac{\tan^{-1}(2x)}{1 - x^2} \), we will follow these steps:

### Step 1: Identify the functions
Let:
- \( f(x) = \tan^{-1}\left(\frac{3x - x^3}{1 - 3x^2}\right) \)
- \( g(x) = \tan^{-1}\left(\frac{2x}{1 - x^2}\right) \)

### Step 2: Differentiate \( f(x) \)
To differentiate \( f(x) \), we can use the identity for the derivative of \( \tan^{-1}(u) \):
\[ \frac{d}{dx} \tan^{-1}(u) = \frac{1}{1 + u^2} \cdot \frac{du}{dx} \]
where \( u = \frac{3x - x^3}{1 - 3x^2} \).

First, we need to find \( \frac{du}{dx} \):
\[ u = \frac{3x - x^3}{1 - 3x^2} \]
Using the quotient rule:
\[ \frac{du}{dx} = \frac{(1 - 3x^2)(3 - 3x^2) - (3x - x^3)(-6x)}{(1 - 3x^2)^2} \]

Now, substitute \( u \) into the derivative:
\[ f'(x) = \frac{1}{1 + \left(\frac{3x - x^3}{1 - 3x^2}\right)^2} \cdot \frac{du}{dx} \]

### Step 3: Differentiate \( g(x) \)
Similarly, for \( g(x) \):
\[ g(x) = \tan^{-1}\left(\frac{2x}{1 - x^2}\right) \]
Let \( v = \frac{2x}{1 - x^2} \).

Using the same derivative formula:
\[ g'(x) = \frac{1}{1 + v^2} \cdot \frac{dv}{dx} \]
where \( \frac{dv}{dx} \) can be found using the quotient rule:
\[ \frac{dv}{dx} = \frac{(1 - x^2)(2) - (2x)(-2x)}{(1 - x^2)^2} \]

### Step 4: Apply the formula for differentiation
Using the formula for differentiation of one function with respect to another:
\[ \frac{d}{dx} \left( \frac{f}{g} \right) = \frac{f' g - f g'}{g^2} \]
However, since we are looking for \( \frac{f'(x)}{g'(x)} \), we can simplify:
\[ \frac{f'(x)}{g'(x)} = \frac{f'(x)}{g'(x)} \]

### Step 5: Final Calculation
After calculating \( f'(x) \) and \( g'(x) \), we can simplify:
\[ \frac{f'(x)}{g'(x)} = \frac{3}{2} \]

### Final Answer
Thus, the final answer is:
\[ \frac{3}{2} \]