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

Options

A. ` (cos x )/(2sqrt sin x) `
B. ` (-cos x)/(2sqrtsin x)`
C. ` (cos x )/(sqrtsin x ) `
D. ` (-cos x)/(sqrtsin x )`

Answer

To find the derivative of the function \( y = \sqrt{\sin x} \), we will use the chain rule and the basic rules of differentiation. Here’s the step-by-step solution:

### Step 1: Rewrite the function
We start with the function:
\[ y = \sqrt{\sin x} \]
This can be rewritten using exponent notation:
\[ y = (\sin x)^{1/2} \]

### Step 2: Differentiate using the chain rule
To differentiate \( y \) with respect to \( x \), we apply the chain rule. The chain rule states that if \( y = u^n \), then:
\[ \frac{dy}{dx} = n \cdot u^{n-1} \cdot \frac{du}{dx} \]
In our case, \( u = \sin x \) and \( n = \frac{1}{2} \). Thus, we have:
\[ \frac{dy}{dx} = \frac{1}{2} (\sin x)^{-1/2} \cdot \frac{d}{dx}(\sin x) \]

### Step 3: Differentiate \( \sin x \)
Now we differentiate \( \sin x \):
\[ \frac{d}{dx}(\sin x) = \cos x \]

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

### Step 5: Rewrite in terms of square root
We can rewrite \( (\sin x)^{-1/2} \) as \( \frac{1}{\sqrt{\sin x}} \):
\[ \frac{dy}{dx} = \frac{\cos x}{2 \sqrt{\sin x}} \]

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

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