If : `y=cos^(-1)(8x^(4)-8x^(2)+1)," then: "(dy)/(dx)=`

Options

A. `(4)/(sqrt(1-x^(2)))`
B. `(-4)/(sqrt(1-x^(2)))`
C. `(4)/(1+x^(2))`
D. `(-4)/(1+x^(2))`

Answer

To find the derivative of the function \( y = \cos^{-1}(8x^4 - 8x^2 + 1) \), we will follow these steps:

### Step 1: Identify the function and apply the chain rule
The function is \( y = \cos^{-1}(u) \) where \( u = 8x^4 - 8x^2 + 1 \).

Using the chain rule, we have:
\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \]

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

### Step 3: Differentiate \( u \) with respect to \( x \)
Now we need to differentiate \( u = 8x^4 - 8x^2 + 1 \):
\[ \frac{du}{dx} = \frac{d}{dx}(8x^4) - \frac{d}{dx}(8x^2) + \frac{d}{dx}(1) \]
Calculating each term:
\[ \frac{d}{dx}(8x^4) = 32x^3, \quad \frac{d}{dx}(8x^2) = 16x, \quad \frac{d}{dx}(1) = 0 \]
Thus,
\[ \frac{du}{dx} = 32x^3 - 16x \]

### Step 4: Substitute \( u \) back into the derivative
Now we can substitute \( u \) back into the derivative:
\[ \frac{dy}{dx} = -\frac{1}{\sqrt{1 - (8x^4 - 8x^2 + 1)^2}} \cdot (32x^3 - 16x) \]

### Step 5: Simplify the expression
We can simplify the expression:
\[ \frac{dy}{dx} = -\frac{32x^3 - 16x}{\sqrt{1 - (8x^4 - 8x^2 + 1)^2}} \]

### Final Result
Thus, the derivative \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = -\frac{16(2x^3 - x)}{\sqrt{1 - (8x^4 - 8x^2 + 1)^2}} \]