Differentiate `tan^(-1) ((3a^2x-x^3)/(a^3-3a x^2)),

Explanation

To differentiate the function \( y = \tan^{-1} \left( \frac{3a^2x - x^3}{a^3 - 3ax^2} \right) \), we will follow these steps:

### Step 1: Rewrite the Function
We start by rewriting the function \( y \) in a more manageable form. We can divide both the numerator and denominator by \( a^3 \):

\[ y = \tan^{-1} \left( \frac{3a^2x - x^3}{a^3 - 3ax^2} \right) = \tan^{-1} \left( \frac{3 \frac{x}{a} - \left( \frac{x}{a} \right)^3}{1 - 3 \left( \frac{x}{a} \right)^2} \right) \]

Let \( u = \frac{x}{a} \), then:

\[ y = \tan^{-1} \left( \frac{3u - u^3}{1 - 3u^2} \right) \]

### Step 2: Use the Formula for \( \tan(3\theta) \)
We recognize that the expression \( \frac{3u - u^3}{1 - 3u^2} \) is the formula for \( \tan(3\theta) \) where \( \theta = \tan^{-1}(u) \):

\[ y = \tan^{-1}(\tan(3\theta)) = 3\theta \]

### Step 3: Substitute Back for \( \theta \)
Since \( \theta = \tan^{-1}\left(\frac{x}{a}\right) \), we can substitute back:

\[ y = 3 \tan^{-1}\left(\frac{x}{a}\right) \]

### Step 4: Differentiate \( y \)
Now we differentiate \( y \) with respect to \( x \):

\[ \frac{dy}{dx} = 3 \cdot \frac{d}{dx} \left( \tan^{-1}\left(\frac{x}{a}\right) \right) \]

Using the derivative formula for \( \tan^{-1}(x) \):

\[ \frac{d}{dx} \left( \tan^{-1}(x) \right) = \frac{1}{1 + x^2} \]

We apply the chain rule:

\[ \frac{d}{dx} \left( \tan^{-1}\left(\frac{x}{a}\right) \right) = \frac{1}{1 + \left(\frac{x}{a}\right)^2} \cdot \frac{d}{dx}\left(\frac{x}{a}\right) = \frac{1}{1 + \frac{x^2}{a^2}} \cdot \frac{1}{a} \]

### Step 5: Simplify the Derivative
Substituting this back into our derivative:

\[ \frac{dy}{dx} = 3 \cdot \frac{1}{1 + \frac{x^2}{a^2}} \cdot \frac{1}{a} = \frac{3}{a} \cdot \frac{1}{\frac{a^2 + x^2}{a^2}} = \frac{3}{a} \cdot \frac{a^2}{a^2 + x^2} \]

Thus, we have:

\[ \frac{dy}{dx} = \frac{3a}{a^2 + x^2} \]

### Final Answer
The derivative of \( y = \tan^{-1} \left( \frac{3a^2x - x^3}{a^3 - 3ax^2} \right) \) is:

\[ \frac{dy}{dx} = \frac{3a}{a^2 + x^2} \]