1. **State the problem:** We are given the function $$f(x) = \frac{1}{2x^2} + x^2 - \sqrt{x} + 5$$ for $$x > 0$$ and need to find its derivative $$f'(x)$$. 2. **Recall derivative rules:** - The derivative of $$x^n$$ is $$nx^{n-1}$$. - The derivative of a constant is 0. - The derivative of $$\sqrt{x} = x^{1/2}$$ is $$\frac{1}{2}x^{-1/2}$$. 3. **Rewrite the function for easier differentiation:** $$f(x) = \frac{1}{2}x^{-2} + x^2 - x^{1/2} + 5$$ 4. **Differentiate each term:** - $$\frac{d}{dx}\left(\frac{1}{2}x^{-2}\right) = \frac{1}{2} \cdot (-2)x^{-3} = -x^{-3} = -\frac{1}{x^3}$$ - $$\frac{d}{dx}(x^2) = 2x$$ - $$\frac{d}{dx}(-x^{1/2}) = -\frac{1}{2}x^{-1/2} = -\frac{1}{2\sqrt{x}}$$ - $$\frac{d}{dx}(5) = 0$$ 5. **Combine the derivatives:** $$f'(x) = -\frac{1}{x^3} + 2x - \frac{1}{2\sqrt{x}}$$ 6. **Check the options:** The correct derivative matches option 3: $$\frac{1}{x^3} + 2x - \frac{1}{2\sqrt{x}}$$ but note the sign of the first term is negative, so the correct derivative is $$-\frac{1}{x^3} + 2x - \frac{1}{2\sqrt{x}}$$. Since the problem's options have $$\frac{1}{x^3}$$ positive, the closest correct derivative is option 3 but with a negative sign on the first term. **Final answer:** $$f'(x) = -\frac{1}{x^3} + 2x - \frac{1}{2\sqrt{x}}$$