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

Answer

To differentiate the function \( y = x^{x^x} \) 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:
\[ \ln y = \ln(x^{x^x}) \]

### Step 2: Simplify using logarithmic properties
Using the property of logarithms that states \( \ln(a^b) = b \ln a \), we can simplify:
\[ \ln y = x^x \ln x \]

### 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:
\[ \frac{d}{dx}(\ln y) = \frac{d}{dx}(x^x \ln x) \]

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

### Step 4: Differentiate \( x^x \)
To differentiate \( x^x \), we can use the logarithmic differentiation again:
\[ x^x = e^{x \ln x} \]
Differentiating \( e^{x \ln x} \) using the chain rule gives:
\[ \frac{d}{dx}(x^x) = x^x (\ln x + 1) \]

### 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 equation
Now we substitute back into our equation:
\[ \frac{1}{y} \frac{dy}{dx} = (x^x (\ln x + 1)) \ln x + x^x \left(\frac{1}{x}\right) \]

### Step 7: Simplify the right side
Combine the terms on the right side:
\[ \frac{1}{y} \frac{dy}{dx} = x^x (\ln x + 1) \ln x + \frac{x^x}{x} \]
\[ \frac{1}{y} \frac{dy}{dx} = x^x (\ln^2 x + \ln x + 1) \]

### Step 8: Solve for \( \frac{dy}{dx} \)
Now, multiply both sides by \( y \):
\[ \frac{dy}{dx} = y \cdot x^x (\ln^2 x + \ln x + 1) \]

### Step 9: Substitute \( y \) back
Recall that \( y = x^{x^x} \):
\[ \frac{dy}{dx} = x^{x^x} \cdot x^x (\ln^2 x + \ln x + 1) \]

### Final Answer
Thus, the derivative of \( y = x^{x^x} \) with respect to \( x \) is:
\[ \frac{dy}{dx} = x^{x^x + x} (\ln^2 x + \ln x + 1) \]

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