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

Answer

To differentiate \( u = \tan^{-1}\left(\frac{3x - x^3}{1 - 3x^2}\right) \) with respect to \( v = \tan^{-1}\left(\frac{x}{\sqrt{1 - x^2}}\right) \), we will follow these steps:

### Step 1: Define the variables
Let:
- \( u = \tan^{-1}\left(\frac{3x - x^3}{1 - 3x^2}\right) \)
- \( v = \tan^{-1}\left(\frac{x}{\sqrt{1 - x^2}}\right) \)

### Step 2: Use the chain rule
We need to find \( \frac{du}{dv} \). By the chain rule, we have:
\[ \frac{du}{dv} = \frac{du/dx}{dv/dx} \]

### Step 3: Differentiate \( u \) with respect to \( x \)
To differentiate \( u \), we apply the derivative of the inverse tangent function:
\[ \frac{du}{dx} = \frac{1}{1 + \left(\frac{3x - x^3}{1 - 3x^2}\right)^2} \cdot \frac{d}{dx}\left(\frac{3x - x^3}{1 - 3x^2}\right) \]

#### Step 3.1: Differentiate the inner function
Let \( y = \frac{3x - x^3}{1 - 3x^2} \). We need to find \( \frac{dy}{dx} \) using the quotient rule:
\[ \frac{dy}{dx} = \frac{(1 - 3x^2)(3 - 3x^2) - (3x - x^3)(-6x)}{(1 - 3x^2)^2} \]

#### Step 3.2: Simplify \( \frac{dy}{dx} \)
Calculating the numerator:
\[ (1 - 3x^2)(3 - 3x^2) + (3x - x^3)(6x) \]
This will yield a polynomial that can be simplified.

### Step 4: Differentiate \( v \) with respect to \( x \)
To differentiate \( v \):
\[ \frac{dv}{dx} = \frac{1}{1 + \left(\frac{x}{\sqrt{1 - x^2}}\right)^2} \cdot \frac{d}{dx}\left(\frac{x}{\sqrt{1 - x^2}}\right) \]

#### Step 4.1: Differentiate the inner function
Let \( z = \frac{x}{\sqrt{1 - x^2}} \). Using the quotient rule:
\[ \frac{dz}{dx} = \frac{\sqrt{1 - x^2} \cdot 1 - x \cdot \frac{-x}{\sqrt{1 - x^2}}}{1 - x^2} \]
This simplifies to:
\[ \frac{dz}{dx} = \frac{1}{(1 - x^2)^{3/2}} \]

### Step 5: Combine results
Now substitute \( \frac{du}{dx} \) and \( \frac{dv}{dx} \) back into the chain rule:
\[ \frac{du}{dv} = \frac{\frac{du}{dx}}{\frac{dv}{dx}} \]

### Step 6: Final expression
After performing all calculations and simplifications, we can express \( \frac{du}{dv} \) in its simplest form.