Differentiate `y=(logx)^(x)`

Answer

To differentiate the function \( y = (\log x)^x \), 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:
\[ \log y = \log((\log x)^x) \]

### Step 2: Apply the logarithmic power rule
Using the property of logarithms that states \( \log(a^b) = b \log a \), we can rewrite the right-hand side:
\[ \log y = x \log(\log x) \]

### Step 3: Differentiate both sides
Now we differentiate both sides with respect to \( x \). For the left-hand side, we use the chain rule:
\[ \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(x \log(\log x)) \]

### Step 4: Apply the product rule on the right-hand side
For the right-hand side, we apply the product rule. Let \( u = x \) and \( v = \log(\log x) \):
\[ \frac{d}{dx}(x \log(\log x)) = u \frac{dv}{dx} + v \frac{du}{dx} \]
Here, \( \frac{du}{dx} = 1 \) and we need to find \( \frac{dv}{dx} \):
\[ \frac{dv}{dx} = \frac{1}{\log x} \cdot \frac{1}{x} = \frac{1}{x \log x} \]
Putting this back into the product rule:
\[ \frac{d}{dx}(x \log(\log x)) = x \cdot \frac{1}{x \log x} + \log(\log x) \cdot 1 = \frac{1}{\log x} + \log(\log x) \]

### Step 5: Substitute back into the equation
Now we can substitute this result back into our differentiated equation:
\[ \frac{1}{y} \frac{dy}{dx} = \frac{1}{\log x} + \log(\log x) \]

### Step 6: Solve for \( \frac{dy}{dx} \)
To isolate \( \frac{dy}{dx} \), we multiply both sides by \( y \):
\[ \frac{dy}{dx} = y \left( \frac{1}{\log x} + \log(\log x) \right) \]

### Step 7: Substitute \( y \) back in
Recall that \( y = (\log x)^x \):
\[ \frac{dy}{dx} = (\log x)^x \left( \frac{1}{\log x} + \log(\log x) \right) \]

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

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