1. **Problem:** Find the second derivative of the function $$f(x) = \sqrt{2x^2 + 3x^{-3} + 7x^{-1}}$$. 2. **Step 1: Rewrite the function** $$f(x) = (2x^2 + 3x^{-3} + 7x^{-1})^{1/2}$$ 3. **Step 2: First derivative using chain rule** Let $$u = 2x^2 + 3x^{-3} + 7x^{-1}$$ Then, $$f'(x) = \frac{1}{2} u^{-1/2} \cdot u'$$ Calculate $$u'$$: $$u' = 4x - 9x^{-4} - 7x^{-2}$$ So, $$f'(x) = \frac{1}{2} (2x^2 + 3x^{-3} + 7x^{-1})^{-1/2} (4x - 9x^{-4} - 7x^{-2})$$ 4. **Step 3: Second derivative using product and chain rules** Let $$v = (2x^2 + 3x^{-3} + 7x^{-1})^{-1/2}$$ and $$w = 4x - 9x^{-4} - 7x^{-2}$$ Then, $$f''(x) = \frac{1}{2} (v' w + v w')$$ Calculate $$v'$$: $$v = u^{-1/2}$$ $$v' = -\frac{1}{2} u^{-3/2} u' = -\frac{1}{2} (2x^2 + 3x^{-3} + 7x^{-1})^{-3/2} (4x - 9x^{-4} - 7x^{-2})$$ Calculate $$w'$$: $$w' = 4 + 36x^{-5} + 14x^{-3}$$ 5. **Step 4: Substitute back and simplify** $$f''(x) = \frac{1}{2} \left[-\frac{1}{2} (2x^2 + 3x^{-3} + 7x^{-1})^{-3/2} (4x - 9x^{-4} - 7x^{-2})^2 + (2x^2 + 3x^{-3} + 7x^{-1})^{-1/2} (4 + 36x^{-5} + 14x^{-3}) \right]$$ 6. **Explanation:** We used the chain rule to differentiate the square root function and product rule for the second derivative. The first derivative is a product of two functions, so the second derivative requires differentiating both parts carefully. **Final answer:** $$f''(x) = \frac{1}{2} \left[-\frac{1}{2} (2x^2 + 3x^{-3} + 7x^{-1})^{-3/2} (4x - 9x^{-4} - 7x^{-2})^2 + (2x^2 + 3x^{-3} + 7x^{-1})^{-1/2} (4 + 36x^{-5} + 14x^{-3}) \right]$$