Differentiate `cos^(-1)(4x^3-3x)` with respect to `x` , if `x in (1/2,\ 1)`

Explanation

To differentiate \( y = \cos^{-1}(4x^3 - 3x) \) with respect to \( x \), we can follow these steps:

### Step 1: Define the function
Let:
\[ y = \cos^{-1}(4x^3 - 3x) \]
where \( x \) is in the interval \( \left( \frac{1}{2}, 1 \right) \).

### Step 2: Use the identity for cosine
Notice that \( 4x^3 - 3x \) can be expressed using the cosine triple angle formula:
\[ \cos(3\theta) = 4\cos^3(\theta) - 3\cos(\theta) \]
If we let \( x = \cos(\theta) \), then:
\[ 4x^3 - 3x = \cos(3\theta) \]
Thus, we can rewrite \( y \) as:
\[ y = \cos^{-1}(\cos(3\theta)) = 3\theta \]

### Step 3: Relate \( \theta \) to \( x \)
Since \( \theta = \cos^{-1}(x) \), we have:
\[ y = 3 \cos^{-1}(x) \]

### Step 4: Differentiate with respect to \( x \)
Now, we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = 3 \cdot \frac{d}{dx}(\cos^{-1}(x)) \]
Using the derivative of \( \cos^{-1}(x) \):
\[ \frac{d}{dx}(\cos^{-1}(x)) = -\frac{1}{\sqrt{1 - x^2}} \]
Thus:
\[ \frac{dy}{dx} = 3 \left(-\frac{1}{\sqrt{1 - x^2}}\right) = -\frac{3}{\sqrt{1 - x^2}} \]

### Final Result
The derivative of \( y = \cos^{-1}(4x^3 - 3x) \) with respect to \( x \) is:
\[ \frac{dy}{dx} = -\frac{3}{\sqrt{1 - x^2}} \]