1. **State the problem:** Given the function $$y(x) = \frac{12x^6 - 4x^4 + 3x^2 - 1}{8x^4}$$, find its derivative $$y'(x)$$ and verify the given derivative $$y'(x) = 3x - \frac{3}{4}x^3 + \frac{1}{2}x^5$$. 2. **Rewrite the function:** To differentiate easily, express $$y(x)$$ as a sum of powers of $$x$$ by dividing each term by $$8x^4$$: $$ y(x) = \frac{12x^6}{8x^4} - \frac{4x^4}{8x^4} + \frac{3x^2}{8x^4} - \frac{1}{8x^4} = \frac{12}{8}x^{6-4} - \frac{4}{8}x^{4-4} + \frac{3}{8}x^{2-4} - \frac{1}{8}x^{-4} $$ Simplify coefficients: $$ y(x) = \frac{3}{2}x^2 - \frac{1}{2} + \frac{3}{8}x^{-2} - \frac{1}{8}x^{-4} $$ 3. **Differentiate term-by-term:** Use the power rule $$\frac{d}{dx} x^n = nx^{n-1}$$: $$ y'(x) = \frac{3}{2} \cdot 2x^{2-1} - 0 + \frac{3}{8} \cdot (-2)x^{-2-1} - \frac{1}{8} \cdot (-4)x^{-4-1} $$ Simplify: $$ y'(x) = 3x - \frac{3}{4}x^{-3} + \frac{1}{2}x^{-5} $$ 4. **Rewrite derivative with positive exponents:** $$ y'(x) = 3x - \frac{3}{4} \frac{1}{x^3} + \frac{1}{2} \frac{1}{x^5} $$ 5. **Compare with given derivative:** The user gave $$y'(x) = 3x - \frac{3}{4}x^3 + \frac{1}{2}x^5$$ which differs in signs of powers. The correct derivative from the function is: $$ y'(x) = 3x - \frac{3}{4}x^{-3} + \frac{1}{2}x^{-5} $$ This means the given derivative is incorrect unless the powers are negative as shown. **Final answer:** $$ y'(x) = 3x - \frac{3}{4}x^{-3} + \frac{1}{2}x^{-5} $$