Differentiate the following w.r.t. x :
`cos^(-1)(4x^(3)-3x),-1ltxlt1`

Answer

To differentiate the function \( y = \cos^{-1}(4x^3 - 3x) \) with respect to \( x \), we will use the chain rule. Here’s the step-by-step solution:

### Step 1: Identify the outer and inner functions
We have:
- Outer function: \( y = \cos^{-1}(u) \) where \( u = 4x^3 - 3x \)

### Step 2: Differentiate the outer function
The derivative of \( \cos^{-1}(u) \) with respect to \( u \) is:
\[ \frac{dy}{du} = -\frac{1}{\sqrt{1 - u^2}} \]

### Step 3: Differentiate the inner function
Now we differentiate \( u = 4x^3 - 3x \) with respect to \( x \):
\[ \frac{du}{dx} = \frac{d}{dx}(4x^3) - \frac{d}{dx}(3x) = 12x^2 - 3 \]

### Step 4: Apply the chain rule
Using the chain rule, we combine the derivatives:
\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = -\frac{1}{\sqrt{1 - (4x^3 - 3x)^2}} \cdot (12x^2 - 3) \]

### Step 5: Simplify the expression
Thus, we can write:
\[ \frac{dy}{dx} = -\frac{12x^2 - 3}{\sqrt{1 - (4x^3 - 3x)^2}} \]

### Final Result
The derivative of \( y = \cos^{-1}(4x^3 - 3x) \) with respect to \( x \) is:
\[ \frac{dy}{dx} = -\frac{12x^2 - 3}{\sqrt{1 - (4x^3 - 3x)^2}} \]

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