`sqrt("log"x)`

Answer

To differentiate the function \( y = \sqrt{\log x} \) with respect to \( x \), we will follow these steps:

### Step 1: Rewrite the function
We can express the square root in terms of a power:
\[ y = (\log x)^{1/2} \]

### Step 2: Apply the chain rule
To differentiate \( y \), we will use the chain rule. The chain rule states that if you have a composite function \( f(g(x)) \), then the derivative is given by:
\[ \frac{dy}{dx} = \frac{dy}{dg} \cdot \frac{dg}{dx} \]
In our case, let \( g(x) = \log x \). Then, we have:
\[ \frac{dy}{dg} = \frac{1}{2} g^{-\frac{1}{2}} = \frac{1}{2 (\log x)^{1/2}} \]

### Step 3: Differentiate \( g(x) \)
Next, we need to find \( \frac{dg}{dx} \):
\[ g(x) = \log x \implies \frac{dg}{dx} = \frac{1}{x} \]

### Step 4: Combine using the chain rule
Now we can combine these results using the chain rule:
\[ \frac{dy}{dx} = \frac{dy}{dg} \cdot \frac{dg}{dx} = \frac{1}{2 (\log x)^{1/2}} \cdot \frac{1}{x} \]

### Step 5: Simplify the expression
Thus, we have:
\[ \frac{dy}{dx} = \frac{1}{2x \sqrt{\log x}} \]

### Final Result
The derivative of \( y = \sqrt{\log x} \) with respect to \( x \) is:
\[ \frac{dy}{dx} = \frac{1}{2x \sqrt{\log x}} \]

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