1. **Problem:** Find the derivative of $$y = \frac{x^3}{3} + \frac{x^2}{2} + \frac{x}{4}$$ **Formula:** Use the power rule $$\frac{d}{dx} x^n = n x^{n-1}$$ **Solution:** $$y' = \frac{d}{dx} \left( \frac{x^3}{3} \right) + \frac{d}{dx} \left( \frac{x^2}{2} \right) + \frac{d}{dx} \left( \frac{x}{4} \right)$$ $$= \frac{3x^{2}}{3} + \frac{2x}{2} + \frac{1}{4} = x^{2} + x + \frac{1}{4}$$ 2. **Problem:** Find the derivative of $$y = -2x^{-1} + \frac{4}{x}$$ **Formula:** Power rule for negative exponents **Solution:** $$y' = -2 \cdot (-1) x^{-2} + 4 \cdot (-1) x^{-2} = 2x^{-2} - 4x^{-2} = -2x^{-2} = -\frac{2}{x^{2}}$$ 3. **Problem:** Find the derivative of $$y = \frac{2x + 5}{3x - 2}$$ **Formula:** Quotient rule $$\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2}$$ **Solution:** Let $$u = 2x + 5, u' = 2$$ Let $$v = 3x - 2, v' = 3$$ $$y' = \frac{2(3x - 2) - (2x + 5)(3)}{(3x - 2)^2} = \frac{6x - 4 - 6x - 15}{(3x - 2)^2} = \frac{-19}{(3x - 2)^2}$$ 4. **Problem:** Find the derivative of $$f(x) = \frac{x^2 - 1}{x^2 + x - 2}$$ **Formula:** Quotient rule **Solution:** $$u = x^2 - 1, u' = 2x$$ $$v = x^2 + x - 2, v' = 2x + 1$$ $$f'(x) = \frac{2x(x^2 + x - 2) - (x^2 - 1)(2x + 1)}{(x^2 + x - 2)^2}$$ Expand numerator: $$2x^3 + 2x^2 - 4x - (2x^3 + x^2 - 2x - 1) = 2x^3 + 2x^2 - 4x - 2x^3 - x^2 + 2x + 1 = x^2 - 2x + 1$$ So, $$f'(x) = \frac{x^2 - 2x + 1}{(x^2 + x - 2)^2} = \frac{(x - 1)^2}{(x^2 + x - 2)^2}$$ 5. **Problem:** Find the derivative of $$f(x) = (1 - x)(1 + x^2)^{-1}$$ **Formula:** Product rule $$\frac{d}{dx}(uv) = u'v + uv'$$ and chain rule for $$v = (1 + x^2)^{-1}$$ **Solution:** $$u = 1 - x, u' = -1$$ $$v = (1 + x^2)^{-1}, v' = -1 \cdot (1 + x^2)^{-2} \cdot 2x = -\frac{2x}{(1 + x^2)^2}$$ $$f'(x) = (-1)(1 + x^2)^{-1} + (1 - x) \left(-\frac{2x}{(1 + x^2)^2}\right)$$ $$= -\frac{1}{1 + x^2} - \frac{2x(1 - x)}{(1 + x^2)^2}$$ 6. **Problem:** Find the derivative of $$r(\theta) = 2 \left( \frac{1}{\sqrt{\theta}} + \sqrt{\theta} \right)$$ **Formula:** Power rule with fractional exponents **Solution:** Rewrite: $$r(\theta) = 2 \left( \theta^{-1/2} + \theta^{1/2} \right)$$ $$r'(\theta) = 2 \left( -\frac{1}{2} \theta^{-3/2} + \frac{1}{2} \theta^{-1/2} \right) = 2 \cdot \frac{1}{2} \left( -\theta^{-3/2} + \theta^{-1/2} \right) = -\theta^{-3/2} + \theta^{-1/2}$$ 7. **Problem:** Find the derivative of $$f(x) = (3 - x^2)(x^3 - x + 1)$$ **Formula:** Product rule **Solution:** $$u = 3 - x^2, u' = -2x$$ $$v = x^3 - x + 1, v' = 3x^2 - 1$$ $$f'(x) = u'v + uv' = (-2x)(x^3 - x + 1) + (3 - x^2)(3x^2 - 1)$$ Expand: $$-2x^4 + 2x^2 - 2x + 9x^2 - 3 - 3x^4 + x^2 = (-2x^4 - 3x^4) + (2x^2 + 9x^2 + x^2) - 2x - 3 = -5x^4 + 12x^2 - 2x - 3$$ 8. **Problem:** Find the derivative of $$f(x) = \frac{(x + 1)(x + 2)}{(x - 1)(x - 2)}$$ **Formula:** Quotient rule **Solution:** $$u = (x + 1)(x + 2) = x^2 + 3x + 2, u' = 2x + 3$$ $$v = (x - 1)(x - 2) = x^2 - 3x + 2, v' = 2x - 3$$ $$f'(x) = \frac{(2x + 3)(x^2 - 3x + 2) - (x^2 + 3x + 2)(2x - 3)}{(x^2 - 3x + 2)^2}$$ Expand numerator: $$(2x + 3)(x^2 - 3x + 2) = 2x^3 - 6x^2 + 4x + 3x^2 - 9x + 6 = 2x^3 - 3x^2 - 5x + 6$$ $$(x^2 + 3x + 2)(2x - 3) = 2x^3 - 3x^2 + 6x^2 - 9x + 4x - 6 = 2x^3 + 3x^2 - 5x - 6$$ Subtract: $$2x^3 - 3x^2 - 5x + 6 - (2x^3 + 3x^2 - 5x - 6) = -6x^2 + 12$$ So, $$f'(x) = \frac{-6x^2 + 12}{(x^2 - 3x + 2)^2} = \frac{-6(x^2 - 2)}{(x^2 - 3x + 2)^2}$$ 9. **Problem:** Find the derivative of $$f(x) = \frac{\sqrt{x} - 1}{\sqrt{x} + 1}$$ **Formula:** Quotient rule **Solution:** $$u = \sqrt{x} - 1 = x^{1/2} - 1, u' = \frac{1}{2} x^{-1/2}$$ $$v = \sqrt{x} + 1 = x^{1/2} + 1, v' = \frac{1}{2} x^{-1/2}$$ $$f'(x) = \frac{\frac{1}{2} x^{-1/2} (x^{1/2} + 1) - (x^{1/2} - 1) \frac{1}{2} x^{-1/2}}{(x^{1/2} + 1)^2}$$ Simplify numerator: $$\frac{1}{2} x^{-1/2} x^{1/2} + \frac{1}{2} x^{-1/2} - \frac{1}{2} x^{-1/2} x^{1/2} + \frac{1}{2} x^{-1/2} = \frac{1}{2} + \frac{1}{2} x^{-1/2} - \frac{1}{2} + \frac{1}{2} x^{-1/2} = x^{-1/2}$$ So, $$f'(x) = \frac{x^{-1/2}}{(\sqrt{x} + 1)^2} = \frac{1}{\sqrt{x} (\sqrt{x} + 1)^2}$$ 10. **Problem:** Find the derivative of $$f(x) = \frac{1}{(x^2 - 1)(x^2 + x + 1)}$$ **Formula:** Use the chain and product rules for denominator and reciprocal rule $$\frac{d}{dx} \frac{1}{g(x)} = -\frac{g'(x)}{g(x)^2}$$ **Solution:** Let $$g(x) = (x^2 - 1)(x^2 + x + 1)$$ $$g'(x) = (2x)(x^2 + x + 1) + (x^2 - 1)(2x + 1)$$ Expand: $$2x^3 + 2x^2 + 2x + 2x^3 + x^2 - 2x - 1 = 4x^3 + 3x^2 - 1$$ So, $$f'(x) = -\frac{4x^3 + 3x^2 - 1}{[(x^2 - 1)(x^2 + x + 1)]^2}$$