`y=sqrt(sinsqrt(x))`

Explanation

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

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

### Step 2: Differentiate Using the Chain Rule
To find \( \frac{dy}{dx} \), we apply the chain rule. The chain rule states that if \( y = f(g(x)) \), then:
\[ \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \]
In our case, let:
- \( f(u) = u^{1/2} \) where \( u = \sin(\sqrt{x}) \)
- \( g(x) = \sqrt{x} \)

Now, we differentiate \( f(u) \) and \( g(x) \):
\[ f'(u) = \frac{1}{2}u^{-1/2} = \frac{1}{2\sqrt{u}} \]
\[ g'(x) = \frac{1}{2\sqrt{x}} \]

### Step 3: Differentiate the Inner Function
Next, we need to differentiate \( \sin(\sqrt{x}) \):
\[ \frac{d}{dx}(\sin(\sqrt{x})) = \cos(\sqrt{x}) \cdot g'(x) = \cos(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}} \]

### Step 4: Combine the Derivatives
Now we can combine everything using the chain rule:
\[ \frac{dy}{dx} = f'(g(x)) \cdot g'(x) = \frac{1}{2\sqrt{\sin(\sqrt{x})}} \cdot \cos(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}} \]

### Step 5: Simplify the Expression
Putting it all together:
\[ \frac{dy}{dx} = \frac{1}{2\sqrt{\sin(\sqrt{x})}} \cdot \cos(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}} = \frac{\cos(\sqrt{x})}{4\sqrt{x}\sqrt{\sin(\sqrt{x})}} \]

### Final Answer
Thus, the derivative of \( y \) with respect to \( x \) is:
\[ \frac{dy}{dx} = \frac{\cos(\sqrt{x})}{4\sqrt{x}\sqrt{\sin(\sqrt{x})}} \]

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