`(2^(x) cotx)/(sqrt(x))`

Answer

To differentiate the function \( y = \frac{2^x \cot x}{\sqrt{x}} \), we will use the quotient rule and product rule of differentiation. Let's go through the steps systematically.

### Step 1: Identify the function
Let \( y = \frac{2^x \cot x}{\sqrt{x}} \).

### Step 2: Apply the quotient rule
The quotient rule states that if you have a function \( y = \frac{u}{v} \), then the derivative \( \frac{dy}{dx} \) is given by:
\[ \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]
Here, let:
- \( u = 2^x \cot x \)
- \( v = \sqrt{x} \)

### Step 3: Differentiate \( u \) and \( v \)
1. **Differentiate \( v \)**:
\[ v = \sqrt{x} = x^{1/2} \implies \frac{dv}{dx} = \frac{1}{2} x^{-1/2} = \frac{1}{2\sqrt{x}} \]

2. **Differentiate \( u \)** using the product rule:
\[ u = 2^x \cot x \]
Let:
- \( a = 2^x \)
- \( b = \cot x \)

Then:
\[ \frac{du}{dx} = a \frac{db}{dx} + b \frac{da}{dx} \]
- \( \frac{db}{dx} = -\csc^2 x \)
- \( \frac{da}{dx} = 2^x \ln 2 \)

Therefore:
\[ \frac{du}{dx} = 2^x (-\csc^2 x) + \cot x (2^x \ln 2) = 2^x \left(-\csc^2 x + \cot x \ln 2\right) \]

### Step 4: Substitute into the quotient rule
Now substitute \( u \), \( v \), \( \frac{du}{dx} \), and \( \frac{dv}{dx} \) into the quotient rule:
\[ \frac{dy}{dx} = \frac{\sqrt{x} \cdot 2^x \left(-\csc^2 x + \cot x \ln 2\right) - 2^x \cot x \cdot \frac{1}{2\sqrt{x}}}{(\sqrt{x})^2} \]

### Step 5: Simplify the expression
The denominator simplifies to:
\[ (\sqrt{x})^2 = x \]
Thus:
\[ \frac{dy}{dx} = \frac{\sqrt{x} \cdot 2^x \left(-\csc^2 x + \cot x \ln 2\right) - \frac{2^x \cot x}{2\sqrt{x}}}{x} \]

### Step 6: Combine terms
Factor out \( 2^x \):
\[ \frac{dy}{dx} = \frac{2^x \left(\sqrt{x} \left(-\csc^2 x + \cot x \ln 2\right) - \frac{\cot x}{2\sqrt{x}}\right)}{x} \]

### Final Result
Thus, the derivative is:
\[ \frac{dy}{dx} = \frac{2^x}{x} \left(\sqrt{x} \left(-\csc^2 x + \cot x \ln 2\right) - \frac{\cot x}{2\sqrt{x}}\right) \]