Differentiate the following w.r.t. x :
`(logx)^(logx),xgt1`

Answer

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

### Step 1: Take the logarithm of both sides
We start by taking the natural logarithm of both sides to simplify the differentiation process:
\[ \log y = \log((\log x)^{\log x}) \]
Using the property of logarithms, we can bring down the exponent:
\[ \log y = \log x \cdot \log(\log x) \]

### Step 2: Differentiate both sides
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:
\[ \frac{d}{dx}(\log y) = \frac{1}{y} \frac{dy}{dx} \]
For the right side, we apply the product rule:
\[ \frac{d}{dx}(\log x \cdot \log(\log x)) = \frac{d}{dx}(\log x) \cdot \log(\log x) + \log x \cdot \frac{d}{dx}(\log(\log x)) \]
Calculating the derivatives:
- The derivative of \( \log x \) is \( \frac{1}{x} \).
- The derivative of \( \log(\log x) \) is \( \frac{1}{\log x} \cdot \frac{1}{x} = \frac{1}{x \log x} \).

Thus, we have:
\[ \frac{d}{dx}(\log x \cdot \log(\log x)) = \frac{1}{x} \cdot \log(\log x) + \log x \cdot \frac{1}{x \log x} = \frac{\log(\log x)}{x} + \frac{1}{x} \]

### Step 3: Set the derivatives equal
Now we set the derivatives equal to each other:
\[ \frac{1}{y} \frac{dy}{dx} = \frac{\log(\log x)}{x} + \frac{1}{x} \]

### Step 4: Solve for \( \frac{dy}{dx} \)
To isolate \( \frac{dy}{dx} \), we multiply both sides by \( y \):
\[ \frac{dy}{dx} = y \left( \frac{\log(\log x)}{x} + \frac{1}{x} \right) \]

### Step 5: Substitute back for \( y \)
Recall that \( y = (\log x)^{\log x} \):
\[ \frac{dy}{dx} = (\log x)^{\log x} \left( \frac{\log(\log x)}{x} + \frac{1}{x} \right) \]

### Final Answer
Thus, the derivative of \( y = (\log x)^{\log x} \) with respect to \( x \) is:
\[ \frac{dy}{dx} = \frac{(\log x)^{\log x}}{x} \left( \log(\log x) + 1 \right) \]