`sin^-1(cos x)+tan ^-1(cot x)`

Answer

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

### Step 1: Differentiate \( y \)
We start with the expression:
\[ y = \sin^{-1}(\cos x) + \tan^{-1}(\cot x) \]

### Step 2: Differentiate \( \sin^{-1}(\cos x) \)
Using the chain rule, the derivative of \( \sin^{-1}(u) \) is given by:
\[ \frac{d}{dx}(\sin^{-1}(u)) = \frac{1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx} \]
Here, \( u = \cos x \), so:
\[ \frac{du}{dx} = -\sin x \]
Thus, we have:
\[ \frac{d}{dx}(\sin^{-1}(\cos x)) = \frac{1}{\sqrt{1 - \cos^2 x}} \cdot (-\sin x) \]
Since \( 1 - \cos^2 x = \sin^2 x \), we can simplify this to:
\[ \frac{d}{dx}(\sin^{-1}(\cos x)) = \frac{-\sin x}{\sqrt{\sin^2 x}} = -\sin x \]

### Step 3: Differentiate \( \tan^{-1}(\cot x) \)
Using the chain rule again, the derivative of \( \tan^{-1}(v) \) is given by:
\[ \frac{d}{dx}(\tan^{-1}(v)) = \frac{1}{1 + v^2} \cdot \frac{dv}{dx} \]
Here, \( v = \cot x \), so:
\[ \frac{dv}{dx} = -\csc^2 x \]
Thus, we have:
\[ \frac{d}{dx}(\tan^{-1}(\cot x)) = \frac{1}{1 + \cot^2 x} \cdot (-\csc^2 x) \]
Since \( 1 + \cot^2 x = \csc^2 x \), we can simplify this to:
\[ \frac{d}{dx}(\tan^{-1}(\cot x)) = \frac{-\csc^2 x}{\csc^2 x} = -1 \]

### Step 4: Combine the derivatives
Now, we can combine the derivatives from Step 2 and Step 3:
\[ \frac{dy}{dx} = -\sin x - 1 \]

### Step 5: Final result
Thus, the derivative of the given expression is:
\[ \frac{dy}{dx} = -\sin x - 1 \]