How do you differentiate x^{1/x}?

To differentiate the function \( y = x^{1/x} \), we can use logarithmic differentiation, which simplifies the process for functions involving variables in both the base and the exponent.

Step 1: Take the Natural Logarithm

Start by taking the natural logarithm of both sides:

Let \( y = x^{1/x} \)

Then, \( \ln(y) = \ln(x^{1/x}) \)

Using the properties of logarithms, this simplifies to:

\( \ln(y) = \frac{1}{x} \ln(x) \)

Step 2: Differentiate Both Sides

Now, differentiate both sides with respect to \( x \):

• For the left side, use the chain rule: \( \frac{1}{y} \frac{dy}{dx} \)
• For the right side, apply the product rule: \( \frac{d}{dx} \left( \frac{1}{x} \ln(x) \right) \)

Calculating the Right Side

Using the product rule:

• Let \( u = \frac{1}{x} \) and \( v = \ln(x) \)
• Then, \( \frac{du}{dx} = -\frac{1}{x^2} \) and \( \frac{dv}{dx} = \frac{1}{x} \)

Applying the product rule gives:

\( \frac{d}{dx} \left( \frac{1}{x} \ln(x) \right) = u \frac{dv}{dx} + v \frac{du}{dx} = \frac{1}{x} \cdot \frac{1}{x} + \ln(x) \cdot \left(-\frac{1}{x^2}\right) \)

This simplifies to:

\( \frac{1}{x^2} - \frac{\ln(x)}{x^2} = \frac{1 - \ln(x)}{x^2} \)

Step 3: Combine Results

Now, equate the derivatives:

\( \frac{1}{y} \frac{dy}{dx} = \frac{1 - \ln(x)}{x^2} \)

To isolate \( \frac{dy}{dx} \), multiply both sides by \( y \):

\( \frac{dy}{dx} = y \cdot \frac{1 - \ln(x)}{x^2} \)

Step 4: Substitute Back for \( y \)

Recall that \( y = x^{1/x} \), so:

\( \frac{dy}{dx} = x^{1/x} \cdot \frac{1 - \ln(x)}{x^2} \)

Final Result

The derivative of \( x^{1/x} \) is:

\( \frac{dy}{dx} = \frac{x^{1/x} (1 - \ln(x))}{x^2} \)