Find `(dy)/(dx)`, when :
`y=((sinx+x^(2)))/(cot 2x)`

Explanation

To find \(\frac{dy}{dx}\) for the function \(y = \frac{\sin x + x^2}{\cot 2x}\), we will use the quotient rule of differentiation. The quotient rule states that if you have a function \(y = \frac{u}{v}\), then the derivative \(\frac{dy}{dx}\) is given by:

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

where \(u = \sin x + x^2\) and \(v = \cot 2x\).

### Step 1: Identify \(u\) and \(v\)
Let:
- \(u = \sin x + x^2\)
- \(v = \cot 2x\)

### Step 2: Differentiate \(u\) and \(v\)
**Differentiate \(u\):**
\[ \frac{du}{dx} = \frac{d}{dx}(\sin x) + \frac{d}{dx}(x^2) = \cos x + 2x \]

**Differentiate \(v\):**
Using the chain rule, we know that \(\frac{d}{dx}(\cot x) = -\csc^2 x\), so:
\[ \frac{dv}{dx} = \frac{d}{dx}(\cot 2x) = -\csc^2(2x) \cdot \frac{d}{dx}(2x) = -2\csc^2(2x) \]

### Step 3: Apply the Quotient Rule
Now we can apply the quotient rule:
\[ \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]

Substituting \(u\), \(v\), \(\frac{du}{dx}\), and \(\frac{dv}{dx}\):
\[ \frac{dy}{dx} = \frac{\cot 2x (\cos x + 2x) - (\sin x + x^2)(-2\csc^2(2x))}{\cot^2(2x)} \]

### Step 4: Simplify the Expression
This can be simplified further:
\[ \frac{dy}{dx} = \frac{\cot 2x (\cos x + 2x) + 2(\sin x + x^2)\csc^2(2x)}{\cot^2(2x)} \]

### Final Result
Thus, the derivative \(\frac{dy}{dx}\) is:
\[ \frac{dy}{dx} = \frac{\cot 2x (\cos x + 2x) + 2(\sin x + x^2)\csc^2(2x)}{\cot^2(2x)} \]