If `y = sqrt(x sin x) " then "(dy)/(dx) =` ?

Options

A. `((x cos x + sin x))/(2 sqrt(x sin x))`
B. `(1)/(2) (x cos x + sin x).sqrt(x sin x)`
C. `(1)/(2 sqrt(x sin x))`
D. none of these

Explanation

To find the derivative of the function \( y = \sqrt{x \sin x} \), we will follow these steps:

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

### Step 2: Apply the chain rule
To differentiate \( y \), we will apply the chain rule. The derivative of \( y = u^{\frac{1}{2}} \) where \( u = x \sin x \) is given by:
\[ \frac{dy}{dx} = \frac{1}{2} u^{-\frac{1}{2}} \cdot \frac{du}{dx} \]
Thus, we have:
\[ \frac{dy}{dx} = \frac{1}{2} (x \sin x)^{-\frac{1}{2}} \cdot \frac{du}{dx} \]

### Step 3: Differentiate \( u = x \sin x \)
Now we need to find \( \frac{du}{dx} \). We will use the product rule for differentiation:
\[ \frac{du}{dx} = \sin x + x \cos x \]
where \( \frac{d}{dx}(x) = 1 \) and \( \frac{d}{dx}(\sin x) = \cos x \).

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

### Step 5: Simplify the expression
We can simplify this expression further:
\[ \frac{dy}{dx} = \frac{\sin x + x \cos x}{2 \sqrt{x \sin x}} \]

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

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