Derivatives of Exponential and Logarithm Functions — Question 8

Question 8 - Solution

We are given: $$f(x) = x^x \quad \text{for } x > 0$$

This is a function where the variable is both the base and the exponent. To differentiate it, we take the natural logarithm of both sides:

Let $$y = x^x$$ Take the natural log of both sides: $$\ln y = \ln(x^x) = x \ln x$$

Differentiate both sides implicitly:

$$\frac{1}{y} \cdot \frac{dy}{dx} = \frac{d}{dx}(x \ln x)$$

Now differentiate the right-hand side using the product rule:

$$\frac{d}{dx}(x \ln x) = 1 \cdot \ln x + x \cdot \frac{1}{x} = \ln x + 1$$

Now solve for $\frac{dy}{dx}$:

$$\frac{dy}{dx} = y(\ln x + 1)$$

Recall that $y = x^x$, so:

$$f'(x) = \boxed{x^x (\ln x + 1)}$$