The derivative of `tan^(-1)((3x-x^(3))/(1-3x^(2)))`w.r.t.x is

Options

A. `(1)/(1+x^(2))`
B. `(3)/(1+9x^(2))`
C. `(3)/(1+x^(2))`
D. `3sec^(2)3x`

Answer

To find the derivative of \( y = \tan^{-1}\left(\frac{3x - x^3}{1 - 3x^2}\right) \) with respect to \( x \), we can follow these steps:

### Step 1: Set up the equation
Let
\[ y = \tan^{-1}\left(\frac{3x - x^3}{1 - 3x^2}\right) \]

### Step 2: Use the tangent identity
We can use the identity for the tangent of a triple angle. Recall that:
\[ \tan(3\theta) = \frac{3\tan(\theta) - \tan^3(\theta)}{1 - 3\tan^2(\theta)} \]
By letting \( x = \tan(\theta) \), we can rewrite:
\[ y = \tan^{-1}(\tan(3\theta)) \]
Thus, we have:
\[ y = 3\theta \]

### Step 3: Relate \( \theta \) to \( x \)
Since \( x = \tan(\theta) \), we can express \( \theta \) in terms of \( x \):
\[ \theta = \tan^{-1}(x) \]
Substituting this back gives:
\[ y = 3\tan^{-1}(x) \]

### Step 4: Differentiate \( y \) with respect to \( x \)
Now, we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = 3 \cdot \frac{d}{dx}(\tan^{-1}(x)) \]
Using the derivative of \( \tan^{-1}(x) \), which is \( \frac{1}{1 + x^2} \), we get:
\[ \frac{dy}{dx} = 3 \cdot \frac{1}{1 + x^2} \]

### Step 5: Final result
Thus, the derivative of \( y \) with respect to \( x \) is:
\[ \frac{dy}{dx} = \frac{3}{1 + x^2} \]