Differentiate the following w.r.t.x :
`sin(cotx)`

Answer

To differentiate the function \( y = \sin(\cot x) \) with respect to \( x \), we will use the chain rule. The chain rule states that if you have a composite function \( y = f(g(x)) \), then the derivative \( \frac{dy}{dx} \) is given by:

\[ \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \]

### Step-by-step Solution:

1. **Identify the outer and inner functions**:
- Outer function: \( f(u) = \sin(u) \) where \( u = \cot x \)
- Inner function: \( g(x) = \cot x \)

2. **Differentiate the outer function**:
- The derivative of \( f(u) = \sin(u) \) is \( f'(u) = \cos(u) \).
- So, \( f'(\cot x) = \cos(\cot x) \).

3. **Differentiate the inner function**:
- The derivative of \( g(x) = \cot x \) is \( g'(x) = -\csc^2(x) \).

4. **Apply the chain rule**:
- Using the chain rule, we have:
\[ \frac{dy}{dx} = f'(\cot x) \cdot g'(x) = \cos(\cot x) \cdot (-\csc^2(x)) \]

5. **Combine the results**:
- Therefore, the derivative is:
\[ \frac{dy}{dx} = -\cos(\cot x) \cdot \csc^2(x) \]

### Final Answer:
\[ \frac{dy}{dx} = -\cos(\cot x) \cdot \csc^2(x) \]