Differentiate the following w.r.t. x :
`(logx)^(x)`

Answer

To differentiate the function \( y = (\log x)^x \) with respect to \( x \), we can follow these steps:

### Step 1: Take the natural logarithm of both sides
We start by taking the natural logarithm of both sides to simplify the differentiation process.

\[ \log y = \log((\log x)^x) \]

### Step 2: Use the properties of logarithms
Using the property of logarithms that states \( \log(a^b) = b \cdot \log a \), we can rewrite the equation:

\[ \log y = x \cdot \log(\log x) \]

### Step 3: Differentiate both sides with respect to \( x \)
Now we differentiate both sides with respect to \( x \). We will use implicit differentiation on the left side and the product rule on the right side.

\[ \frac{d}{dx}(\log y) = \frac{1}{y} \cdot \frac{dy}{dx} \]

For the right side, we apply the product rule:

\[ \frac{d}{dx}(x \cdot \log(\log x)) = \log(\log x) + x \cdot \frac{d}{dx}(\log(\log x)) \]

### Step 4: Differentiate \( \log(\log x) \)
To differentiate \( \log(\log x) \), we use the chain rule:

\[ \frac{d}{dx}(\log(\log x)) = \frac{1}{\log x} \cdot \frac{1}{x} = \frac{1}{x \log x} \]

### Step 5: Substitute back into the differentiation equation
Now we substitute this back into our differentiation equation:

\[ \frac{1}{y} \cdot \frac{dy}{dx} = \log(\log x) + x \cdot \frac{1}{x \log x} \]

This simplifies to:

\[ \frac{1}{y} \cdot \frac{dy}{dx} = \log(\log x) + \frac{1}{\log x} \]

### Step 6: Solve for \( \frac{dy}{dx} \)
Now, we multiply both sides by \( y \) to solve for \( \frac{dy}{dx} \):

\[ \frac{dy}{dx} = y \left( \log(\log x) + \frac{1}{\log x} \right) \]

### Step 7: Substitute back for \( y \)
Since \( y = (\log x)^x \), we substitute back:

\[ \frac{dy}{dx} = (\log x)^x \left( \log(\log x) + \frac{1}{\log x} \right) \]

### Final Answer
Thus, the derivative of \( y = (\log x)^x \) with respect to \( x \) is:

\[ \frac{dy}{dx} = (\log x)^x \left( \log(\log x) + \frac{1}{\log x} \right) \]

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