If `y=sin^(-1) (2 cos^(2)x-1),"then " dy/dx=`

Options

A. `1`
B. `-1`
C. `2`
D. `-2`

Answer

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

### Step 1: Rewrite the function using a trigonometric identity
We know that:
\[ 2 \cos^2 x - 1 = \cos(2x) \]
Thus, we can rewrite \( y \) as:
\[ y = \sin^{-1}(\cos(2x)) \]

### Step 2: Differentiate using the chain rule
To differentiate \( y \), we will use the formula for the derivative of the inverse sine function:
\[ \frac{dy}{dx} = \frac{1}{\sqrt{1 - (\cos(2x))^2}} \cdot \frac{d}{dx}(\cos(2x)) \]

### Step 3: Differentiate \( \cos(2x) \)
The derivative of \( \cos(2x) \) is:
\[ \frac{d}{dx}(\cos(2x)) = -\sin(2x) \cdot 2 = -2\sin(2x) \]

### Step 4: Substitute back into the derivative formula
Now substituting back, we have:
\[ \frac{dy}{dx} = \frac{1}{\sqrt{1 - \cos^2(2x)}} \cdot (-2\sin(2x)) \]

### Step 5: Simplify the expression
Using the identity \( 1 - \cos^2(\theta) = \sin^2(\theta) \), we can simplify:
\[ \sqrt{1 - \cos^2(2x)} = \sin(2x) \]
Thus, we have:
\[ \frac{dy}{dx} = \frac{-2\sin(2x)}{\sin(2x)} = -2 \]

### Final Answer
Therefore, the derivative is:
\[ \frac{dy}{dx} = -2 \]

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