Differentiate the following w.r.t. x :
`cos^(-1)(sinx)`

Answer

To differentiate \( y = \cos^{-1}(\sin x) \) with respect to \( x \), we will use the chain rule of differentiation. Here’s a step-by-step solution:

### Step 1: Identify the outer and inner functions
Let \( u = \sin x \). Then we can rewrite \( y \) as:
\[ y = \cos^{-1}(u) \]

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

### Step 3: Differentiate the inner function
Next, we differentiate \( u = \sin x \) with respect to \( x \):
\[ \frac{du}{dx} = \cos x \]

### Step 4: Apply the chain rule
Now, we apply the chain rule:
\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \]
Substituting the derivatives we found:
\[ \frac{dy}{dx} = -\frac{1}{\sqrt{1 - u^2}} \cdot \cos x \]

### Step 5: Substitute back for \( u \)
Since \( u = \sin x \), we substitute back:
\[ \frac{dy}{dx} = -\frac{1}{\sqrt{1 - \sin^2 x}} \cdot \cos x \]

### Step 6: Simplify the expression
Using the identity \( \cos^2 x + \sin^2 x = 1 \), we have:
\[ 1 - \sin^2 x = \cos^2 x \]
Thus, the square root becomes:
\[ \sqrt{1 - \sin^2 x} = \cos x \]
Now substituting this into our derivative:
\[ \frac{dy}{dx} = -\frac{1}{\cos x} \cdot \cos x \]
This simplifies to:
\[ \frac{dy}{dx} = -1 \]

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