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

Explanation

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

### Step 1: Substitute \( x \) with \( a \tan(\theta) \)
Let \( x = a \tan(\theta) \). Then, we can express \( y \) in terms of \( \theta \).

### Step 2: Rewrite the function
Substituting \( x = a \tan(\theta) \) into the function gives:
\[ y = \tan^{-1}\left(\frac{3a^2(a \tan(\theta)) - (a \tan(\theta))^3}{a^3 - 3a(a \tan(\theta))^2}\right) \]
This simplifies to:
\[ y = \tan^{-1}\left(\frac{3a^3 \tan(\theta) - a^3 \tan^3(\theta)}{a^3 - 3a^3 \tan^2(\theta)}\right) \]

### Step 3: Simplify the expression
Factoring out \( a^3 \) from the numerator and denominator:
\[ y = \tan^{-1}\left(\frac{a^3(3 \tan(\theta) - \tan^3(\theta))}{a^3(1 - 3 \tan^2(\theta))}\right) \]
This simplifies to:
\[ y = \tan^{-1}\left(\frac{3 \tan(\theta) - \tan^3(\theta)}{1 - 3 \tan^2(\theta)}\right) \]

### Step 4: Use the tangent triple angle formula
We know that:
\[ \tan(3\theta) = \frac{3 \tan(\theta) - \tan^3(\theta)}{1 - 3 \tan^2(\theta)} \]
Thus, we can rewrite \( y \):
\[ y = \tan^{-1}(\tan(3\theta)) = 3\theta \]

### Step 5: Substitute back for \( \theta \)
Since \( \theta = \tan^{-1}\left(\frac{x}{a}\right) \), we have:
\[ y = 3 \tan^{-1}\left(\frac{x}{a}\right) \]

### Step 6: Differentiate with respect to \( x \)
Now we differentiate \( y \):
\[ \frac{dy}{dx} = 3 \cdot \frac{d}{dx}\left(\tan^{-1}\left(\frac{x}{a}\right)\right) \]
Using the derivative of \( \tan^{-1}(u) \):
\[ \frac{d}{dx}\left(\tan^{-1}(u)\right) = \frac{1}{1 + u^2} \cdot \frac{du}{dx} \]
where \( u = \frac{x}{a} \), thus \( \frac{du}{dx} = \frac{1}{a} \).

### Step 7: Final differentiation
Now substituting back:
\[ \frac{dy}{dx} = 3 \cdot \frac{1}{1 + \left(\frac{x}{a}\right)^2} \cdot \frac{1}{a} \]
This simplifies to:
\[ \frac{dy}{dx} = \frac{3}{a(1 + \frac{x^2}{a^2})} = \frac{3}{a\left(\frac{a^2 + x^2}{a^2}\right)} = \frac{3a}{a^2 + x^2} \]

### Final Answer:
\[ \frac{dy}{dx} = \frac{3a}{a^2 + x^2} \]