Find the maximum value of the function `x^(1//x)`.

Answer

To find the maximum value of the function \( y = x^{1/x} \), we will follow these steps:

### Step 1: Take the natural logarithm of both sides
We start by taking the natural logarithm of the function:
\[ \log y = \log(x^{1/x}) \]
Using the property of logarithms, we can simplify this:
\[ \log y = \frac{1}{x} \log x \]

### Step 2: Differentiate both sides
Next, we differentiate both sides with respect to \( x \). Using implicit differentiation:
\[ \frac{d}{dx}(\log y) = \frac{1}{y} \frac{dy}{dx} \]
For the right side, we use the quotient rule:
\[ \frac{d}{dx}\left(\frac{\log x}{x}\right) = \frac{x \cdot \frac{1}{x} - \log x \cdot 1}{x^2} = \frac{1 - \log x}{x^2} \]
Thus, we have:
\[ \frac{1}{y} \frac{dy}{dx} = \frac{1 - \log x}{x^2} \]

### Step 3: Solve for \( \frac{dy}{dx} \)
Multiplying both sides by \( y \):
\[ \frac{dy}{dx} = y \cdot \frac{1 - \log x}{x^2} \]
Substituting back \( y = x^{1/x} \):
\[ \frac{dy}{dx} = x^{1/x} \cdot \frac{1 - \log x}{x^2} \]

### Step 4: Set the derivative to zero
To find the critical points, we set \( \frac{dy}{dx} = 0 \):
\[ x^{1/x} \cdot \frac{1 - \log x}{x^2} = 0 \]
Since \( x^{1/x} \) is never zero for \( x > 0 \), we focus on:
\[ 1 - \log x = 0 \implies \log x = 1 \implies x = e \]

### Step 5: Determine if it is a maximum or minimum
To confirm whether this critical point is a maximum or minimum, we can find the second derivative or analyze the sign of the first derivative around \( x = e \). For simplicity, we can evaluate the first derivative around \( e \):
- For \( x < e \), \( 1 - \log x > 0 \) (positive).
- For \( x > e \), \( 1 - \log x < 0 \) (negative).

This indicates that \( y \) increases on \( (0, e) \) and decreases on \( (e, \infty) \), confirming that \( x = e \) is a maximum.

### Step 6: Calculate the maximum value
Finally, we calculate the maximum value of the function at \( x = e \):
\[ y = e^{1/e} \]

Thus, the maximum value of the function \( x^{1/x} \) is:
\[ \boxed{e^{1/e}} \]