1. **Problem 1a:** Differentiate $$f(x) = \frac{3x^2 - 2}{4x^{-3} - 2\sqrt{4\sqrt{x} + 1}}$$ - Use the quotient rule: $$\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$$ where $$u = 3x^2 - 2$$ and $$v = 4x^{-3} - 2\sqrt{4\sqrt{x} + 1}$$. - Compute $$u' = 6x$$. - For $$v'$$, differentiate each term: - $$\frac{d}{dx}(4x^{-3}) = 4 \cdot (-3)x^{-4} = -12x^{-4}$$. - For $$-2\sqrt{4\sqrt{x} + 1}$$, rewrite as $$-2(4x^{1/2} + 1)^{1/2}$$. - Use chain rule: $$\frac{d}{dx} (4x^{1/2} + 1)^{1/2} = \frac{1}{2}(4x^{1/2} + 1)^{-1/2} \cdot 4 \cdot \frac{1}{2} x^{-1/2} = \frac{1}{2} (4x^{1/2} + 1)^{-1/2} \cdot 2 x^{-1/2} = (4x^{1/2} + 1)^{-1/2} x^{-1/2}$$. - So $$v' = -12x^{-4} - 2 \cdot (4x^{1/2} + 1)^{-1/2} x^{-1/2}$$. - Apply quotient rule: $$f'(x) = \frac{6x (4x^{-3} - 2\sqrt{4\sqrt{x} + 1}) - (3x^2 - 2)(-12x^{-4} - 2 (4x^{1/2} + 1)^{-1/2} x^{-1/2})}{(4x^{-3} - 2\sqrt{4\sqrt{x} + 1})^2}$$. 2. **Problem 1b:** Differentiate $$f(x) = e^x (x^3 - 4x + 1) + \frac{\ln(x^2 - 4)}{x^2 - 4}$$ - Use product rule for $$e^x (x^3 - 4x + 1)$$: $$\frac{d}{dx}[e^x (x^3 - 4x + 1)] = e^x (x^3 - 4x + 1) + e^x (3x^2 - 4)$$. - For $$\frac{\ln(x^2 - 4)}{x^2 - 4}$$, use quotient rule with $$u = \ln(x^2 - 4)$$ and $$v = x^2 - 4$$: - $$u' = \frac{2x}{x^2 - 4}$$ - $$v' = 2x$$ - Quotient rule: $$\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} = \frac{\frac{2x}{x^2 - 4} (x^2 - 4) - \ln(x^2 - 4) (2x)}{(x^2 - 4)^2} = \frac{2x - 2x \ln(x^2 - 4)}{(x^2 - 4)^2}$$. - Combine all: $$f'(x) = e^x (x^3 - 4x + 1) + e^x (3x^2 - 4) + \frac{2x - 2x \ln(x^2 - 4)}{(x^2 - 4)^2}$$. **Final answers:** $$f'(x) = \frac{6x (4x^{-3} - 2\sqrt{4\sqrt{x} + 1}) - (3x^2 - 2)(-12x^{-4} - 2 (4x^{1/2} + 1)^{-1/2} x^{-1/2})}{(4x^{-3} - 2\sqrt{4\sqrt{x} + 1})^2}$$ $$f'(x) = e^x (x^3 - 4x + 1) + e^x (3x^2 - 4) + \frac{2x - 2x \ln(x^2 - 4)}{(x^2 - 4)^2}$$