1. Problem: Find the derivative of $$y = (2x^2 - 4x^3)^4$$. Step 1: Use the chain rule. Let $$u = 2x^2 - 4x^3$$, so $$y = u^4$$. Step 2: Compute $$\frac{dy}{du} = 4u^3$$. Step 3: Compute $$\frac{du}{dx} = 4x - 12x^2$$. Step 4: Apply the chain rule: $$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = 4(2x^2 - 4x^3)^3 (4x - 12x^2)$$. 2. Problem: Find all existing higher derivatives of $$y = x^5 - 2x^4 + x^3 - 3x^2 - 4x + 7$$. Step 1: First derivative: $$y' = 5x^4 - 8x^3 + 3x^2 - 6x - 4$$ Step 2: Second derivative: $$y'' = 20x^3 - 24x^2 + 6x - 6$$ Step 3: Third derivative: $$y''' = 60x^2 - 48x + 6$$ Step 4: Fourth derivative: $$y^{(4)} = 120x - 48$$ Step 5: Fifth derivative: $$y^{(5)} = 120$$ Step 6: Sixth derivative and higher derivatives: $$y^{(6)}= 0$$ and all higher derivatives are zero. 3. Problem: Find the derivative of $$y = \sqrt{x^2 + 1}$$. Step 1: Rewrite: $$y = (x^2 + 1)^{1/2}$$. Step 2: Use the chain rule: $$\frac{dy}{dx} = \frac{1}{2}(x^2 + 1)^{-1/2} \cdot 2x = \frac{x}{\sqrt{x^2 + 1}}$$. 4. Problem: Given $$y = \frac{u^2}{u^2 + 1}$$ and $$u = \sqrt{2x + 1}$$. (i) Find $$\frac{dy}{dx}$$ by the chain rule. Step 1: Rewrite $$y$$ as $$y = \frac{u^2}{u^2 + 1}$$. Step 2: Compute $$\frac{dy}{du}$$ using quotient rule: $$\frac{dy}{du} = \frac{(2u)(u^2 + 1) - u^2(2u)}{(u^2 + 1)^2} = \frac{2u^3 + 2u - 2u^3}{(u^2 + 1)^2} = \frac{2u}{(u^2 + 1)^2}$$. Step 3: Compute $$\frac{du}{dx}$$: $$u = (2x + 1)^{1/2}, \quad \frac{du}{dx} = \frac{1}{2}(2x+1)^{-1/2} \cdot 2 = \frac{1}{\sqrt{2x + 1}}$$. Step 4: Chain rule: $$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{2u}{(u^2 + 1)^2} \cdot \frac{1}{\sqrt{2x + 1}}$$. Step 5: Substitute back $$u = \sqrt{2x + 1}$$: $$\frac{dy}{dx} = \frac{2 \sqrt{2x + 1}}{((2x + 1) + 1)^2} \cdot \frac{1}{\sqrt{2x + 1}} = \frac{2}{(2x + 2)^2} = \frac{2}{4(x+1)^2} = \frac{1}{2(x+1)^2}$$. (ii) Construct the composite function and then find $$\frac{dy}{dx}$$. Step 1: Substitute $$u = \sqrt{2x + 1}$$ into $$y$$: $$y = \frac{(\sqrt{2x + 1})^2}{(\sqrt{2x + 1})^2 + 1} = \frac{2x + 1}{2x + 1 + 1} = \frac{2x + 1}{2x + 2}$$. Step 2: Simplify denominator: $$y = \frac{2x + 1}{2(x + 1)}$$. Step 3: Find derivative using quotient rule: $$\frac{dy}{dx} = \frac{2 \cdot 2(x+1) - (2x + 1) \cdot 2}{(2(x+1))^2} = \frac{4(x+1) - 2(2x + 1)}{4(x+1)^2} = \frac{4x + 4 - 4x - 2}{4(x+1)^2} = \frac{2}{4(x+1)^2} = \frac{1}{2(x+1)^2}$$. Step 4: Result matches the chain rule method. 5. Problem: Find $$\frac{dy}{dx}$$ if $$y = \sin^3(4x^2 + 2x)$$. Step 1: Rewrite as $$y = (\sin(4x^2 + 2x))^3$$. Step 2: Use chain rule: $$\frac{dy}{dx} = 3(\sin(4x^2 + 2x))^2 \cdot \cos(4x^2 + 2x) \cdot \frac{d}{dx}(4x^2 + 2x)$$. Step 3: Compute inner derivative: $$\frac{d}{dx}(4x^2 + 2x) = 8x + 2$$. Step 4: Final derivative: $$\frac{dy}{dx} = 3 \sin^2(4x^2 + 2x) \cdot \cos(4x^2 + 2x) \cdot (8x + 2)$$. 6. Problem: Find the derivative of $$y = \tan 4x^2$$. Step 1: Use chain rule: $$\frac{dy}{dx} = \sec^2(4x^2) \cdot \frac{d}{dx}(4x^2) = \sec^2(4x^2) \cdot 8x$$. 7. Problem: Find $$\frac{dy}{dx}$$ if $$y = \sin(\ln x)$$. Step 1: Use chain rule: $$\frac{dy}{dx} = \cos(\ln x) \cdot \frac{d}{dx}(\ln x) = \cos(\ln x) \cdot \frac{1}{x} = \frac{\cos(\ln x)}{x}$$. 8. Problem: Find $$\frac{dy}{dx}$$ if $$y = \frac{\sin x}{x}$$. Step 1: Use quotient rule: $$\frac{dy}{dx} = \frac{x \cdot \cos x - \sin x \cdot 1}{x^2} = \frac{x \cos x - \sin x}{x^2}$$. 9. Problem: Find the increments. (i) $$F(x) = 2x^2 + 3x - 5, x=2, \Delta x = 0.5$$. Step 1: Compute $$F(2) = 2(2)^2 + 3(2) - 5 = 8 + 6 - 5 = 9$$. Step 2: Compute $$F(2.5) = 2(2.5)^2 + 3(2.5) - 5 = 2(6.25) + 7.5 - 5 = 12.5 + 7.5 - 5 = 15$$. Step 3: Increment $$\Delta F = F(2.5) - F(2) = 15 - 9 = 6$$. (ii) $$g(x) = \frac{x^2 - 4}{x - 2}, x=1, \Delta x = 2$$. Step 1: Compute $$g(1) = \frac{1 - 4}{1 - 2} = \frac{-3}{-1} = 3$$. Step 2: Compute $$g(3) = \frac{9 - 4}{3 - 2} = \frac{5}{1} = 5$$. Step 3: Increment $$\Delta g = g(3) - g(1) = 5 - 3 = 2$$.