1. **Problem statement:** Find the derivative of the function $$f(x) = \frac{e^x}{x^2}$$ for $$x > 0$$. 2. **Formula used:** We will use the quotient rule for derivatives, which states: $$\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$$ where $$u = e^x$$ and $$v = x^2$$. 3. **Calculate derivatives of numerator and denominator:** - $$u' = \frac{d}{dx} e^x = e^x$$ - $$v' = \frac{d}{dx} x^2 = 2x$$ 4. **Apply the quotient rule:** $$f'(x) = \frac{e^x \cdot x^2 - e^x \cdot 2x}{(x^2)^2} = \frac{e^x x^2 - 2x e^x}{x^4}$$ 5. **Factor the numerator:** $$f'(x) = \frac{e^x (x^2 - 2x)}{x^4} = \frac{e^x x (x - 2)}{x^4}$$ 6. **Simplify the expression:** $$f'(x) = \frac{e^x (x - 2)}{x^3}$$ **Final answer:** $$\boxed{f'(x) = \frac{e^x (x - 2)}{x^3}}$$