`sqrt(x sinx)`

Answer

To differentiate the function \( y = \sqrt{x \sin x} \) with respect to \( x \), we will use the chain rule and the product rule. Here’s a step-by-step solution:

### Step 1: Rewrite the function
We start by rewriting the function in a more convenient form for differentiation:
\[ y = (x \sin x)^{1/2} \]

### Step 2: Apply the chain rule
Using the chain rule, we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = \frac{1}{2}(x \sin x)^{-1/2} \cdot \frac{d}{dx}(x \sin x) \]

### Step 3: Differentiate \( x \sin x \) using the product rule
Now we need to differentiate \( x \sin x \). We apply the product rule, which states that if \( u = x \) and \( v = \sin x \), then:
\[ \frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx} \]
Here, \( \frac{du}{dx} = 1 \) and \( \frac{dv}{dx} = \cos x \). Therefore:
\[ \frac{d}{dx}(x \sin x) = x \cos x + \sin x \]

### Step 4: Substitute back into the derivative
Now we substitute this result back into our expression for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{1}{2}(x \sin x)^{-1/2} \cdot (x \cos x + \sin x) \]

### Step 5: Rewrite the final expression
We can rewrite the final expression for the derivative:
\[ \frac{dy}{dx} = \frac{x \cos x + \sin x}{2 \sqrt{x \sin x}} \]

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

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