1. **State the problem:** Differentiate the function $$f(x) = \frac{(x^2 + 1)^3}{x}$$ with respect to $x$. 2. **Rewrite the function:** To differentiate more easily, express $f(x)$ as a product: $$f(x) = (x^2 + 1)^3 \cdot x^{-1}$$ 3. **Identify the rule to use:** Since $f(x)$ is a product of two functions, use the **product rule**: $$\frac{d}{dx}[u \cdot v] = u'v + uv'$$ where $u = (x^2 + 1)^3$ and $v = x^{-1}$. 4. **Differentiate $u$:** Use the **chain rule** for $u = (x^2 + 1)^3$: $$u' = 3(x^2 + 1)^2 \cdot \frac{d}{dx}(x^2 + 1) = 3(x^2 + 1)^2 \cdot 2x = 6x(x^2 + 1)^2$$ 5. **Differentiate $v$:** $$v = x^{-1} \implies v' = -1 \cdot x^{-2} = -x^{-2}$$ 6. **Apply the product rule:** $$f'(x) = u'v + uv' = 6x(x^2 + 1)^2 \cdot x^{-1} + (x^2 + 1)^3 \cdot (-x^{-2})$$ 7. **Simplify each term:** $$6x(x^2 + 1)^2 \cdot x^{-1} = 6(x^2 + 1)^2$$ $$ (x^2 + 1)^3 \cdot (-x^{-2}) = -\frac{(x^2 + 1)^3}{x^2}$$ 8. **Combine terms:** $$f'(x) = 6(x^2 + 1)^2 - \frac{(x^2 + 1)^3}{x^2}$$ 9. **Express with common denominator $x^2$:** $$f'(x) = \frac{6(x^2 + 1)^2 \cdot x^2}{x^2} - \frac{(x^2 + 1)^3}{x^2} = \frac{6x^2(x^2 + 1)^2 - (x^2 + 1)^3}{x^2}$$ 10. **Factor numerator:** $$= \frac{(x^2 + 1)^2 \left(6x^2 - (x^2 + 1)\right)}{x^2} = \frac{(x^2 + 1)^2 (6x^2 - x^2 - 1)}{x^2} = \frac{(x^2 + 1)^2 (5x^2 - 1)}{x^2}$$ **Final answer:** $$f'(x) = \frac{(x^2 + 1)^2 (5x^2 - 1)}{x^2}$$