`sin^(-1)(3x-4x^(3))`

Answer

To differentiate the function \( y = \sin^{-1}(3x - 4x^3) \), we will follow these steps:

### Step 1: Identify the function
We have:
\[ y = \sin^{-1}(3x - 4x^3) \]

### Step 2: Differentiate using the chain rule
To differentiate \( y \) with respect to \( x \), we will use the chain rule. The derivative of \( \sin^{-1}(u) \) is given by:
\[ \frac{dy}{dx} = \frac{1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx} \]
where \( u = 3x - 4x^3 \).

### Step 3: Find \( \frac{du}{dx} \)
Now, we need to find \( \frac{du}{dx} \):
\[ u = 3x - 4x^3 \]
Differentiating \( u \) with respect to \( x \):
\[ \frac{du}{dx} = 3 - 12x^2 \]

### Step 4: Substitute \( u \) and \( \frac{du}{dx} \) into the chain rule formula
Now we substitute \( u \) and \( \frac{du}{dx} \) into the chain rule formula:
\[ \frac{dy}{dx} = \frac{1}{\sqrt{1 - (3x - 4x^3)^2}} \cdot (3 - 12x^2) \]

### Step 5: Simplify the expression
The expression can be simplified, but we will leave it in this form for now:
\[ \frac{dy}{dx} = \frac{3 - 12x^2}{\sqrt{1 - (3x - 4x^3)^2}} \]

### Final Result
Thus, the derivative of \( y = \sin^{-1}(3x - 4x^3) \) is:
\[ \frac{dy}{dx} = \frac{3 - 12x^2}{\sqrt{1 - (3x - 4x^3)^2}} \]