1. Problem a: Find the derivative of $$y=\frac{x \ln x}{1-x^2}$$ using the quotient rule. 2. Let $$u = x \ln x$$ and $$v = 1 - x^2$$. Then: - $$u' = 1 \cdot \ln x + x \cdot \frac{1}{x} = \ln x + 1$$ by the product rule. - $$v' = -2x$$. 3. By the quotient rule: $$y' = \frac{u'v - uv'}{v^2} = \frac{(\ln x + 1)(1-x^2) - (x \ln x)(-2x)}{(1-x^2)^2}$$ 4. Simplify numerator: $$ (\ln x +1)(1-x^2) + 2x^2 \ln x = (\ln x +1)(1-x^2) + 2 x^2 \ln x $$ $$= (\ln x +1) - (\ln x +1) x^2 + 2 x^2 \ln x$$ $$= \ln x + 1 - x^2 \ln x - x^2 + 2 x^2 \ln x$$ $$= \ln x + 1 - x^2 + x^2 \ln x$$ 5. Final derivative for a: $$y' = \frac{\ln x + 1 - x^2 + x^2 \ln x}{(1 - x^2)^2}$$ 6. Problem b: Find derivative of $$y = 2^x x^2$$ using product rule. 7. Let $$u = 2^x$$ and $$v = x^2$$. - $$u' = 2^x \ln 2$$ by exponential derivative rule. - $$v' = 2x$$. 8. By product rule: $$y' = u' v + u v' = 2^x \ln 2 \cdot x^2 + 2^x \cdot 2x = 2^x x^2 \ln 2 + 2^{x+1} x$$ 9. Problem c: Derivative of $$y = \frac{9}{2} \sqrt[3]{x^2} - 16 \sqrt{x^5} - 14$$. 10. Rewrite terms with exponents: - $$\sqrt[3]{x^2} = x^{2/3}$$ - $$\sqrt{x^5} = x^{5/2}$$ 11. Derivative: $$y' = \frac{9}{2} \cdot \frac{2}{3} x^{-1/3} - 16 \cdot \frac{5}{2} x^{3/2} - 0 = 3 x^{-1/3} - 40 x^{3/2}$$ 12. Final answers: - a) $$y' = \frac{\ln x + 1 - x^2 + x^2 \ln x}{(1 - x^2)^2}$$ - b) $$y' = 2^x x^2 \ln 2 + 2^{x+1} x$$ - c) $$y' = 3 x^{-1/3} - 40 x^{3/2}$$