Dfferentiate `sin^-1(cosx)` with respect to x.

Options

A. 0
B. -1
C. 2
D. 1

Answer

To differentiate \( y = \sin^{-1}(\cos x) \) with respect to \( x \), we will follow these steps:

### Step 1: Define the function
Let
\[ y = \sin^{-1}(\cos x) \]

### Step 2: Differentiate using the chain rule
To differentiate \( y \) with respect to \( x \), we use the chain rule. The derivative of \( \sin^{-1}(u) \) with respect to \( u \) is given by:
\[ \frac{d}{du}(\sin^{-1}(u)) = \frac{1}{\sqrt{1 - u^2}} \]
where \( u = \cos x \). Thus, we have:
\[ \frac{dy}{dx} = \frac{1}{\sqrt{1 - (\cos x)^2}} \cdot \frac{d}{dx}(\cos x) \]

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

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

### Step 5: Simplify the expression
We know that \( 1 - \cos^2 x = \sin^2 x \). Therefore, we can simplify:
\[ \frac{dy}{dx} = \frac{-\sin x}{\sqrt{\sin^2 x}} = \frac{-\sin x}{|\sin x|} \]

### Step 6: Determine the sign of \( \sin x \)
The expression \( \frac{-\sin x}{|\sin x|} \) can be simplified further:
- If \( \sin x > 0 \), then \( |\sin x| = \sin x \) and \( \frac{-\sin x}{\sin x} = -1 \).
- If \( \sin x < 0 \), then \( |\sin x| = -\sin x \) and \( \frac{-\sin x}{-\sin x} = 1 \).
- If \( \sin x = 0 \), the expression is undefined.

Thus, we can conclude:
\[ \frac{dy}{dx} = \begin{cases} -1 & \text{if } \sin x > 0 \\ 1 & \text{if } \sin x < 0 \\ \text{undefined} & \text{if } \sin x = 0 \end{cases} \]

### Final Answer
The derivative of \( \sin^{-1}(\cos x) \) with respect to \( x \) is:
\[ \frac{dy}{dx} = \frac{-\sin x}{|\sin x|} \]