`Fi nd dy/dx , y=tan^-1[(3x-x^3)/(1-3x^2)],-1/sqrt3ltxlt1/sqrt3`

Explanation

To find \(\frac{dy}{dx}\) for the function \(y = \tan^{-1}\left(\frac{3x - x^3}{1 - 3x^2}\right)\), we can use a known identity for the tangent inverse function.

### Step-by-Step Solution:

1. **Recognize the Identity**:
We know from trigonometric identities that:
\[ \tan^{-1}\left(\frac{3x - x^3}{1 - 3x^2}\right) = 3\tan^{-1}(x) \]
Therefore, we can rewrite \(y\) as:
\[ y = 3\tan^{-1}(x) \]

2. **Differentiate Both Sides**:
Now, we differentiate both sides with respect to \(x\):
\[ \frac{dy}{dx} = 3 \cdot \frac{d}{dx}(\tan^{-1}(x)) \]

3. **Use the Derivative of \(\tan^{-1}(x)\)**:
The derivative of \(\tan^{-1}(x)\) is:
\[ \frac{d}{dx}(\tan^{-1}(x)) = \frac{1}{1 + x^2} \]
Substituting this into our equation gives:
\[ \frac{dy}{dx} = 3 \cdot \frac{1}{1 + x^2} \]

4. **Final Result**:
Thus, we have:
\[ \frac{dy}{dx} = \frac{3}{1 + x^2} \]