1. **Problem:** Find the derivative of $$y = \frac{x^3}{3} + \frac{x^2}{2} + \frac{x}{4}$$ Formula: $$\frac{d}{dx} x^n = n x^{n-1}$$ Step 1: Differentiate each term: $$\frac{d}{dx} \left( \frac{x^3}{3} \right) = \frac{3x^{2}}{3} = x^2$$ $$\frac{d}{dx} \left( \frac{x^2}{2} \right) = \frac{2x}{2} = x$$ $$\frac{d}{dx} \left( \frac{x}{4} \right) = \frac{1}{4}$$ Step 2: Combine results: $$y' = x^2 + x + \frac{1}{4}$$ --- 2. **Problem:** Find the derivative of $$y = -2x^{-1} + \frac{4}{x}$$ Rewrite $$\frac{4}{x} = 4x^{-1}$$ Step 1: Differentiate each term: $$\frac{d}{dx} (-2x^{-1}) = -2 \cdot (-1) x^{-2} = 2x^{-2}$$ $$\frac{d}{dx} (4x^{-1}) = 4 \cdot (-1) x^{-2} = -4x^{-2}$$ Step 2: Combine results: $$y' = 2x^{-2} - 4x^{-2} = -2x^{-2} = -\frac{2}{x^2}$$ --- 3. **Problem:** Find the derivative of $$y = \frac{2x + 5}{3x - 2}$$ Formula: Quotient rule $$\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v u' - u v'}{v^2}$$ Let $$u = 2x + 5, u' = 2$$ $$v = 3x - 2, v' = 3$$ Step 1: Apply quotient rule: $$y' = \frac{(3x - 2)(2) - (2x + 5)(3)}{(3x - 2)^2}$$ Step 2: Expand numerator: $$(3x - 2)(2) = 6x - 4$$ $$(2x + 5)(3) = 6x + 15$$ Step 3: Subtract: $$6x - 4 - (6x + 15) = 6x - 4 - 6x - 15 = -19$$ Step 4: Final derivative: $$y' = \frac{-19}{(3x - 2)^2}$$ --- 4. **Problem:** Find the derivative of $$f(x) = \frac{x^2 - 1}{x^2 + x - 2}$$ Factor denominator: $$x^2 + x - 2 = (x + 2)(x - 1)$$ Let $$u = x^2 - 1, u' = 2x$$ $$v = x^2 + x - 2, v' = 2x + 1$$ Step 1: Quotient rule: $$f'(x) = \frac{v u' - u v'}{v^2} = \frac{(x^2 + x - 2)(2x) - (x^2 - 1)(2x + 1)}{(x^2 + x - 2)^2}$$ Step 2: Expand numerator: $$(x^2 + x - 2)(2x) = 2x^3 + 2x^2 - 4x$$ $$(x^2 - 1)(2x + 1) = 2x^3 + x^2 - 2x - 1$$ Step 3: Subtract: $$2x^3 + 2x^2 - 4x - (2x^3 + x^2 - 2x - 1) = 2x^3 + 2x^2 - 4x - 2x^3 - x^2 + 2x + 1 = x^2 - 2x + 1$$ Step 4: Final derivative: $$f'(x) = \frac{x^2 - 2x + 1}{(x^2 + x - 2)^2} = \frac{(x - 1)^2}{(x^2 + x - 2)^2}$$ --- 5. **Problem:** Find the derivative of $$f(x) = (1 - x)(1 + x^2)^{-1}$$ Use product rule: $$f'(x) = u'v + uv'$$ Let $$u = 1 - x, u' = -1$$ $$v = (1 + x^2)^{-1}, v' = -1 \cdot (1 + x^2)^{-2} \cdot 2x = -\frac{2x}{(1 + x^2)^2}$$ Step 1: Apply product rule: $$f'(x) = (-1)(1 + x^2)^{-1} + (1 - x) \left(-\frac{2x}{(1 + x^2)^2}\right)$$ Step 2: Simplify: $$f'(x) = -\frac{1}{1 + x^2} - \frac{2x(1 - x)}{(1 + x^2)^2}$$ Step 3: Combine over common denominator: $$f'(x) = -\frac{(1 + x^2)}{(1 + x^2)^2} - \frac{2x(1 - x)}{(1 + x^2)^2} = \frac{-(1 + x^2) - 2x(1 - x)}{(1 + x^2)^2}$$ Step 4: Expand numerator: $$-(1 + x^2) - 2x + 2x^2 = -1 - x^2 - 2x + 2x^2 = -1 + x^2 - 2x$$ Step 5: Final derivative: $$f'(x) = \frac{x^2 - 2x - 1}{(1 + x^2)^2}$$ --- 6. **Problem:** Find the derivative of $$r(\theta) = 2 \left( \frac{1}{\sqrt{\theta}} + \sqrt{\theta} \right)$$ Rewrite: $$r(\theta) = 2 \left( \theta^{-1/2} + \theta^{1/2} \right)$$ Step 1: Differentiate inside parentheses: $$\frac{d}{d\theta} \theta^{-1/2} = -\frac{1}{2} \theta^{-3/2}$$ $$\frac{d}{d\theta} \theta^{1/2} = \frac{1}{2} \theta^{-1/2}$$ Step 2: Multiply by 2: $$r'(\theta) = 2 \left(-\frac{1}{2} \theta^{-3/2} + \frac{1}{2} \theta^{-1/2} \right) = -\theta^{-3/2} + \theta^{-1/2}$$ Step 3: Final derivative: $$r'(\theta) = \frac{1}{\sqrt{\theta}} - \frac{1}{\theta^{3/2}}$$ --- 7. **Problem:** Find the derivative of $$f(x) = (3 - x^2)(x^3 - x + 1)$$ Use product rule: $$f'(x) = u'v + uv'$$ Let $$u = 3 - x^2, u' = -2x$$ $$v = x^3 - x + 1, v' = 3x^2 - 1$$ Step 1: Apply product rule: $$f'(x) = (-2x)(x^3 - x + 1) + (3 - x^2)(3x^2 - 1)$$ Step 2: Expand terms: $$-2x(x^3 - x + 1) = -2x^4 + 2x^2 - 2x$$ $$(3 - x^2)(3x^2 - 1) = 9x^2 - 3 - 3x^4 + x^2 = -3x^4 + 10x^2 - 3$$ Step 3: Combine: $$f'(x) = (-2x^4 + 2x^2 - 2x) + (-3x^4 + 10x^2 - 3) = -5x^4 + 12x^2 - 2x - 3$$ --- 8. **Problem:** Find the derivative of $$f(x) = \frac{(x + 1)(x + 2)}{(x - 1)(x - 2)}$$ Let $$u = (x + 1)(x + 2) = x^2 + 3x + 2, u' = 2x + 3$$ $$v = (x - 1)(x - 2) = x^2 - 3x + 2, v' = 2x - 3$$ Step 1: Quotient rule: $$f'(x) = \frac{v u' - u v'}{v^2} = \frac{(x^2 - 3x + 2)(2x + 3) - (x^2 + 3x + 2)(2x - 3)}{(x^2 - 3x + 2)^2}$$ Step 2: Expand numerator: $$(x^2 - 3x + 2)(2x + 3) = 2x^3 + 3x^2 - 6x^2 - 9x + 4x + 6 = 2x^3 - 3x^2 - 5x + 6$$ $$(x^2 + 3x + 2)(2x - 3) = 2x^3 - 3x^2 + 6x^2 - 9x + 4x - 6 = 2x^3 + 3x^2 - 5x - 6$$ Step 3: Subtract: $$(2x^3 - 3x^2 - 5x + 6) - (2x^3 + 3x^2 - 5x - 6) = -6x^2 + 12$$ Step 4: Final derivative: $$f'(x) = \frac{-6x^2 + 12}{(x^2 - 3x + 2)^2} = \frac{-6(x^2 - 2)}{(x^2 - 3x + 2)^2}$$ --- 9. **Problem:** Find the derivative of $$f(x) = \frac{\sqrt{x} - 1}{\sqrt{x} + 1}$$ Rewrite $$\sqrt{x} = x^{1/2}$$ Let $$u = x^{1/2} - 1, u' = \frac{1}{2} x^{-1/2}$$ $$v = x^{1/2} + 1, v' = \frac{1}{2} x^{-1/2}$$ Step 1: Quotient rule: $$f'(x) = \frac{v u' - u v'}{v^2} = \frac{(x^{1/2} + 1) \frac{1}{2} x^{-1/2} - (x^{1/2} - 1) \frac{1}{2} x^{-1/2}}{(x^{1/2} + 1)^2}$$ Step 2: Factor $$\frac{1}{2} x^{-1/2}$$: $$f'(x) = \frac{\frac{1}{2} x^{-1/2} \left( (x^{1/2} + 1) - (x^{1/2} - 1) \right)}{(x^{1/2} + 1)^2} = \frac{\frac{1}{2} x^{-1/2} (2)}{(x^{1/2} + 1)^2}$$ Step 3: Simplify numerator: $$f'(x) = \frac{x^{-1/2}}{(\sqrt{x} + 1)^2} = \frac{1}{\sqrt{x} (\sqrt{x} + 1)^2}$$ --- 10. **Problem:** Find the derivative of $$f(x) = \frac{1}{(x^2 - 1)(x^2 + x + 1)}$$ Rewrite: $$f(x) = \left[(x^2 - 1)(x^2 + x + 1)\right]^{-1}$$ Let $$g(x) = (x^2 - 1)(x^2 + x + 1)$$ Step 1: Use chain rule: $$f'(x) = -1 \cdot g(x)^{-2} \cdot g'(x) = -\frac{g'(x)}{g(x)^2}$$ Step 2: Find $$g'(x)$$ using product rule: $$g'(x) = (x^2 - 1)' (x^2 + x + 1) + (x^2 - 1)(x^2 + x + 1)'$$ Calculate derivatives: $$(x^2 - 1)' = 2x$$ $$(x^2 + x + 1)' = 2x + 1$$ Step 3: Substitute: $$g'(x) = 2x (x^2 + x + 1) + (x^2 - 1)(2x + 1)$$ Step 4: Expand: $$2x (x^2 + x + 1) = 2x^3 + 2x^2 + 2x$$ $$(x^2 - 1)(2x + 1) = 2x^3 + x^2 - 2x - 1$$ Step 5: Sum: $$g'(x) = (2x^3 + 2x^2 + 2x) + (2x^3 + x^2 - 2x - 1) = 4x^3 + 3x^2 - 1$$ Step 6: Final derivative: $$f'(x) = -\frac{4x^3 + 3x^2 - 1}{\left[(x^2 - 1)(x^2 + x + 1)\right]^2}$$