`y=logsqrt(sinsqrt(e^(x)))`

Explanation

To differentiate the function \( y = \log(\sqrt{\sin(\sqrt{e^x})}) \), we will follow these steps:

### Step 1: Rewrite the function
We can rewrite the function using properties of logarithms and exponents:
\[ y = \log(\sqrt{\sin(\sqrt{e^x})}) = \frac{1}{2} \log(\sin(\sqrt{e^x})) \]

### Step 2: Differentiate using the chain rule
Now, we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{\sin(\sqrt{e^x})} \cdot \frac{d}{dx}(\sin(\sqrt{e^x})) \]

### Step 3: Differentiate \( \sin(\sqrt{e^x}) \)
Next, we need to differentiate \( \sin(\sqrt{e^x}) \) using the chain rule:
\[ \frac{d}{dx}(\sin(\sqrt{e^x})) = \cos(\sqrt{e^x}) \cdot \frac{d}{dx}(\sqrt{e^x}) \]

### Step 4: Differentiate \( \sqrt{e^x} \)
Now, we differentiate \( \sqrt{e^x} \):
\[ \frac{d}{dx}(\sqrt{e^x}) = \frac{1}{2\sqrt{e^x}} \cdot e^x \]

### Step 5: Combine the derivatives
Now we can substitute back into our expression:
\[ \frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{\sin(\sqrt{e^x})} \cdot \left( \cos(\sqrt{e^x}) \cdot \frac{1}{2\sqrt{e^x}} \cdot e^x \right) \]

### Step 6: Simplify the expression
Now we simplify:
\[ \frac{dy}{dx} = \frac{1}{4} \cdot \frac{e^x \cos(\sqrt{e^x})}{\sin(\sqrt{e^x}) \sqrt{e^x}} \]
This can be written as:
\[ \frac{dy}{dx} = \frac{1}{4} \cdot \frac{e^x \cos(\sqrt{e^x})}{\sin(\sqrt{e^x}) \sqrt{e^x}} = \frac{1}{4} \cdot \frac{e^x}{\sqrt{e^x}} \cdot \cot(\sqrt{e^x}) \]

### Final Result
Thus, the final result is:
\[ \frac{dy}{dx} = \frac{e^{x/2} \cot(\sqrt{e^x})}{4} \]