Differentiate `(sinx)/((1+cotx))`

Answer

To differentiate the function \( y = \frac{\sin x}{1 + \cot x} \), we will use the quotient rule. The quotient rule states that if you have a function in the form \( \frac{u}{v} \), then the derivative is given by:

\[ \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]

where \( u = \sin x \) and \( v = 1 + \cot x \).

### Step 1: Identify \( u \) and \( v \)

Let:
- \( u = \sin x \)
- \( v = 1 + \cot x \)

### Step 2: Differentiate \( u \) and \( v \)

Now, we need to find \( \frac{du}{dx} \) and \( \frac{dv}{dx} \).

- \( \frac{du}{dx} = \cos x \)
- To differentiate \( v = 1 + \cot x \), we use the derivative of \( \cot x \):
\[ \frac{dv}{dx} = 0 - \csc^2 x = -\csc^2 x \]

### Step 3: Apply the Quotient Rule

Now, we can apply the quotient rule:

\[ \frac{dy}{dx} = \frac{(1 + \cot x)(\cos x) - (\sin x)(-\csc^2 x)}{(1 + \cot x)^2} \]

### Step 4: Simplify the Expression

Now, let's simplify the numerator:

1. The first term is \( (1 + \cot x) \cos x \).
2. The second term becomes \( \sin x \csc^2 x = \sin x \cdot \frac{1}{\sin^2 x} = \frac{1}{\sin x} \).

So, the numerator becomes:

\[ (1 + \cot x)\cos x + \frac{1}{\sin x} \]

Now, substituting back into the derivative:

\[ \frac{dy}{dx} = \frac{(1 + \cot x)\cos x + \csc^2 x}{(1 + \cot x)^2} \]

### Final Answer

Thus, the derivative of \( y = \frac{\sin x}{1 + \cot x} \) is:

\[ \frac{dy}{dx} = \frac{(1 + \cot x)\cos x + \csc^2 x}{(1 + \cot x)^2} \]