If ` y = sin^(-1) (cos x) ` , then derivative of y is

Options

A. `-1`
B. 0
C. 1
D. `(1)/(2)`

Explanation

To find the derivative of \( y = \sin^{-1}(\cos x) \), we will use the chain rule and the derivative of the inverse sine function. Here’s how to solve it step by step:

### Step 1: Differentiate \( y \) with respect to \( x \)

Given:
\[ y = \sin^{-1}(\cos x) \]

Using the chain rule, the derivative of \( y \) with respect to \( x \) is:
\[ \frac{dy}{dx} = \frac{d}{dx}(\sin^{-1}(u)) \cdot \frac{du}{dx} \]
where \( u = \cos x \).

### Step 2: Find \( \frac{du}{dx} \)

Now, we need to find the derivative of \( u = \cos x \):
\[ \frac{du}{dx} = -\sin x \]

### Step 3: Use the derivative of \( \sin^{-1}(u) \)

The derivative of \( \sin^{-1}(u) \) with respect to \( u \) is:
\[ \frac{d}{du}(\sin^{-1}(u)) = \frac{1}{\sqrt{1 - u^2}} \]

### Step 4: Substitute \( u \) back into the derivative

Now, substituting \( u = \cos x \):
\[ \frac{dy}{dx} = \frac{1}{\sqrt{1 - \cos^2 x}} \cdot (-\sin x) \]

### Step 5: Simplify the expression

Using the Pythagorean identity \( \sin^2 x + \cos^2 x = 1 \), we can simplify \( 1 - \cos^2 x \) to \( \sin^2 x \):
\[ \frac{dy}{dx} = \frac{-\sin x}{\sqrt{\sin^2 x}} \]

Since \( \sqrt{\sin^2 x} = |\sin x| \), we can write:
\[ \frac{dy}{dx} = \frac{-\sin x}{|\sin x|} \]

### Step 6: Determine the sign

This expression simplifies to:
\[ \frac{dy}{dx} = -1 \quad \text{if } \sin x > 0 \]
\[ \frac{dy}{dx} = 1 \quad \text{if } \sin x < 0 \]

However, if we are looking for the derivative in general, we can state:
\[ \frac{dy}{dx} = -1 \quad \text{(for intervals where } \sin x \neq 0\text{)} \]

### Final Answer:
Thus, the derivative of \( y = \sin^{-1}(\cos x) \) is:
\[ \frac{dy}{dx} = -1 \]

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