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

Answer

To differentiate the function \( y = x^{(x^x)} \) with respect to \( x \), we will 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.

\[ \ln y = \ln(x^{(x^x)}) \]

### Step 2: Apply the logarithmic identity
Using the property of logarithms, we can bring down the exponent:

\[ \ln y = x^x \cdot \ln x \]

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

\[ \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(x^x \ln x) \]

Using the product rule on the right side:

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

### Step 4: Differentiate \( x^x \)
To differentiate \( x^x \), we can use the logarithmic differentiation again:

\[ x^x = e^{x \ln x} \]
Thus,
\[ \frac{d}{dx}(x^x) = x^x \left( \ln x + 1 \right) \]

### Step 5: Differentiate \( \ln x \)
The derivative of \( \ln x \) is:

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

### Step 6: Substitute back into the differentiation
Now substituting back into our equation:

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

### Step 7: Simplify the right side
Combine the terms on the right side:

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

### Step 8: Solve for \( \frac{dy}{dx} \)
Now, multiply both sides by \( y \):

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

### Step 9: Substitute back for \( y \)
Recall that \( y = x^{(x^x)} \):

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

### Final Result
Thus, the derivative of \( y = x^{(x^x)} \) with respect to \( x \) is:

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

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