Find `(dy)/(dx)`, when:
`y=(logx)^(x)`

Answer

To find \(\frac{dy}{dx}\) when \(y = (\log x)^x\), we can follow these steps:

### Step 1: Take the natural logarithm of both sides
We start by taking the logarithm of both sides to simplify the differentiation process:
\[ \log y = \log((\log x)^x) \]

### Step 2: Apply the logarithmic property
Using the property of logarithms, we can simplify the right-hand side:
\[ \log y = x \cdot \log(\log x) \]

### Step 3: 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:
\[ \frac{d}{dx}(\log y) = \frac{d}{dx}(x \cdot \log(\log x)) \]

Using the chain rule on the left side:
\[ \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(x) \cdot \log(\log x) + x \cdot \frac{d}{dx}(\log(\log x)) \]

### Step 4: Differentiate the right-hand side
Now we differentiate the right-hand side:
1. The derivative of \(x\) is \(1\).
2. The derivative of \(\log(\log x)\) using the chain rule is:
\[ \frac{1}{\log x} \cdot \frac{1}{x} = \frac{1}{x \log x} \]

Putting it all together:
\[ \frac{1}{y} \frac{dy}{dx} = \log(\log x) + x \cdot \frac{1}{x \log x} \]
This simplifies to:
\[ \frac{1}{y} \frac{dy}{dx} = \log(\log x) + \frac{1}{\log x} \]

### Step 5: Solve for \(\frac{dy}{dx}\)
Now we multiply both sides by \(y\) to isolate \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = y \left(\log(\log x) + \frac{1}{\log x}\right) \]

### Step 6: Substitute back for \(y\)
Recall that \(y = (\log x)^x\):
\[ \frac{dy}{dx} = (\log x)^x \left(\log(\log x) + \frac{1}{\log x}\right) \]

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

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