1. **State the problem:** Calculate the first derivative of the function $$f(x) = \frac{x^2 - x}{x^2}$$. 2. **Rewrite the function:** Simplify the function before differentiating. $$f(x) = \frac{x^2 - x}{x^2} = \frac{x^2}{x^2} - \frac{x}{x^2} = 1 - \frac{1}{x}$$ 3. **Recall the derivative rules:** - The derivative of a constant is 0. - The derivative of $$x^n$$ is $$nx^{n-1}$$. - The derivative of $$\frac{1}{x} = x^{-1}$$ is $$-x^{-2} = -\frac{1}{x^2}$$. 4. **Differentiate term by term:** $$f'(x) = 0 - \left(-\frac{1}{x^2}\right) = \frac{1}{x^2}$$ 5. **Final answer:** $$f'(x) = \frac{1}{x^2}$$ 6. **Check options:** None of the options exactly match $$\frac{1}{x^2}$$, but let's verify by differentiating the original function using the quotient rule to confirm. 7. **Using quotient rule:** For $$f(x) = \frac{u}{v}$$, $$f'(x) = \frac{u'v - uv'}{v^2}$$ where $$u = x^2 - x$$ and $$v = x^2$$. Calculate derivatives: $$u' = 2x - 1$$ $$v' = 2x$$ Apply quotient rule: $$f'(x) = \frac{(2x - 1)(x^2) - (x^2 - x)(2x)}{(x^2)^2} = \frac{(2x - 1)x^2 - 2x(x^2 - x)}{x^4}$$ Simplify numerator: $$= \frac{2x^3 - x^2 - 2x^3 + 2x^2}{x^4} = \frac{(-x^2 + 2x^2)}{x^4} = \frac{x^2}{x^4}$$ Simplify fraction: $$= \frac{\cancel{x^2}}{x^4} = \frac{1}{x^{4-2}} = \frac{1}{x^2}$$ 8. **Conclusion:** The derivative is $$f'(x) = \frac{1}{x^2}$$ which matches none of the options given exactly. **Therefore, the correct derivative is $$\boxed{\frac{1}{x^2}}$$, which is not listed among the options A) to D).