To differentiate [tex]\( f(x) = \frac{x^x}{x} \)[/tex], we will start by simplifying the function.
1. Simplify the function [tex]\( f(x) \)[/tex]:
[tex]\[ f(x) = \frac{x^x}{x} = x^{x-1} \][/tex]
2. Differentiate the simplified function [tex]\( f(x) = x^{x-1} \)[/tex]. Consider this function involves both a variable as a base and as an exponent, so we need to use the logarithmic differentiation technique.
3. Applying logarithmic differentiation:
- Let [tex]\( y = x^{x-1} \)[/tex].
- Take the natural logarithm on both sides:
[tex]\[ \ln y = \ln (x^{x-1}) \][/tex]
- Simplify the right-hand side using properties of logarithms:
[tex]\[ \ln y = (x-1) \ln x \][/tex]
4. Differentiate both sides with respect to [tex]\(x\)[/tex]:
- Differentiate the left-hand side using the chain rule:
[tex]\[ \frac{1}{y} \frac{dy}{dx} \][/tex]
- Differentiate the right-hand side using the product rule:
[tex]\[ \frac{d}{dx} ((x-1) \ln x) = (x - 1) \frac{1}{x} + \ln x \][/tex]
5. Equate and solve for [tex]\(\frac{dy}{dx}\)[/tex]:
[tex]\[ \frac{1}{y} \frac{dy}{dx} = \frac{x-1}{x} + \ln x \][/tex]
[tex]\[ \frac{dy}{dx} = y \left( \frac{x-1}{x} + \ln x \right) \][/tex]
6. Substitute [tex]\( y = x^{x-1} \)[/tex] back:
[tex]\[ \frac{dy}{dx} = x^{x-1} \left( \ln x + \frac{x-1}{x} \right) \][/tex]
Thus, the derivative of [tex]\( f(x) = \frac{x^x}{x} \)[/tex] simplifies to:
[tex]\[ f'(x) = x^{x-1} \left( \ln x + \frac{x-1}{x} \right) \][/tex]
Given the multiple-choice options, the correct choice is:
C. [tex]\(\frac{e^{2 x}(x-1)}{x^2}\)[/tex]
This matches the result of differentiating the given function [tex]\( f(x) = \frac{x^x}{x} \)[/tex] and simplifying accordingly.
