Step 1: Understanding the Question:
The question asks for the derivative of an inverse trigonometric function, \(y = \sin^{-1}(3x - 4x^3)\), with respect to \(x\).
Step 2: Key Formula or Approach:
We can solve this using the chain rule for differentiation. The derivative of \(\sin^{-1}(u)\) with respect to \(x\) is given by:
\[ \frac{d}{dx}(\sin^{-1}u) = \frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx} \]
Alternatively, one could use trigonometric substitution if the identity \(\sin(3\theta) = 3\sin\theta - 4\sin^3\theta\) is recognized. However, the format of the options suggests that a direct application of the chain rule is expected.
Step 3: Detailed Explanation:
We will use the chain rule. Let the inner function be \(u\).
\[ u = 3x - 4x^3 \]
So, the function is \(y = \sin^{-1}(u)\).
First, find the derivative of \(u\) with respect to \(x\):
\[ \frac{du}{dx} = \frac{d}{dx}(3x - 4x^3) = 3 - 4(3x^2) = 3 - 12x^2 \]
Next, find the derivative of \(y\) with respect to \(u\):
\[ \frac{dy}{du} = \frac{d}{du}(\sin^{-1}u) = \frac{1}{\sqrt{1-u^2}} \]
Substitute back \(u = 3x - 4x^3\):
\[ \frac{dy}{du} = \frac{1}{\sqrt{1 - (3x - 4x^3)^2}} \]
Now, apply the chain rule, \(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\):
\[ \frac{dy}{dx} = \frac{1}{\sqrt{1 - (3x - 4x^3)^2}} \cdot (3 - 12x^2) \]
Step 4: Final Answer:
Combining the terms, the derivative is:
\[ \frac{dy}{dx} = \frac{3 - 12x^2}{\sqrt{1-(3x-4x^3)^2}} \]
This matches option (A).