If `y=cos(2cos^(-1)x)," then "y''=`

Options

A. 3
B. 4
C. 1
D. 0

Answer

To solve the problem, we need to find the second derivative \( y'' \) of the function \( y = \cos(2 \cos^{-1}(x)) \).

### Step-by-Step Solution:

1. **Rewrite the Function Using an Identity**:
We start with the function:
\[ y = \cos(2 \cos^{-1}(x)) \]
We can use the trigonometric identity:
\[ \cos(2\theta) = 2\cos^2(\theta) - 1 \]
Here, let \( \theta = \cos^{-1}(x) \). Thus, we can rewrite \( y \) as:
\[ y = 2 \cos^2(\cos^{-1}(x)) - 1 \]

2. **Simplify Using the Inverse Function**:
Since \( \cos(\cos^{-1}(x)) = x \), we have:
\[ \cos^2(\cos^{-1}(x)) = x^2 \]
Therefore, substituting this back into the equation for \( y \):
\[ y = 2x^2 - 1 \]

3. **Find the First Derivative**:
Now we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = \frac{d}{dx}(2x^2 - 1) \]
The derivative of \( 2x^2 \) is \( 4x \) and the derivative of \( -1 \) is \( 0 \):
\[ \frac{dy}{dx} = 4x \]

4. **Find the Second Derivative**:
Next, we differentiate \( \frac{dy}{dx} \) to find \( y'' \):
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}(4x) \]
The derivative of \( 4x \) is \( 4 \):
\[ y'' = 4 \]

### Final Answer:
Thus, the second derivative \( y'' \) is:
\[ \boxed{4} \]