1. Find \( \frac{d y}{d x} \) for each of the following functions. \( \begin{array}{lll}\text { (a) } y=4 x^{2}(5 x+3) & \text { (b) } y=-2 x^{3}(x+1) & \text { (c) } y=x^{2}(1-4 x)^{4} \\ \text { (d) } y=x^{2} \sqrt{1-2 x^{2}} & \text { (e) } y=(4 x-3)(2 x+7)^{6} & \text { (f) } y=(x+5)^{3}(x-4)^{4}\end{array} \)

To find \( \frac{d y}{d x} \) for each of the given functions, we can apply the product rule and the chain rule of differentiation. Let's break it down! 1. **For (a)**: \( y = 4x^2(5x + 3) \) Using the product rule: \( \frac{d y}{d x} = 4x^2 \cdot 5 + (5x + 3) \cdot 8x = 20x^2 + 8x(5x + 3) = 20x^2 + 40x^2 + 24x = 60x^2 + 24x. \) 2. **For (b)**: \( y = -2x^3(x + 1) \) Applying the product rule: \( \frac{d y}{d x} = -2x^3 \cdot 1 + (x + 1) \cdot -6x^2 = -2x^3 - 6x^2(x + 1) = -2x^3 - 6x^3 - 6x^2 = -8x^3 - 6x^2. \) 3. **For (c)**: \( y = x^2(1 - 4x)^4 \) Using the product rule and chain rule: \( \frac{d y}{d x} = x^2 \cdot 4(1 - 4x)^3 \cdot (-4) + (1 - 4x)^4 \cdot 2x = -16x^2(1 - 4x)^3 + 2x(1 - 4x)^4. \) 4. **For (d)**: \( y = x^2 \sqrt{1 - 2x^2} \) Product rule and chain rule yield: \( \frac{d y}{d x} = x^2 \cdot \frac{1}{2\sqrt{1 - 2x^2}} \cdot (-4x) + 2x \sqrt{1 - 2x^2} = -2x^3/\sqrt{1 - 2x^2} + 2x \sqrt{1 - 2x^2.} \) 5. **For (e)**: \( y = (4x - 3)(2x + 7)^6 \) Using product and chain rule: \( \frac{d y}{d x} = (4)(2x + 7)^6 + (4x - 3)(6(2x + 7)^5 \cdot 2) = 4(2x + 7)^6 + 12(4x - 3)(2x + 7)^5. \) 6. **For (f)**: \( y = (x + 5)^3(x - 4)^4 \) Using product rule: \( \frac{d y}{d x} = 3(x + 5)^2(x - 4)^4 + (x + 5)^3 \cdot 4(x - 4)^3. \) And there you have it! The derivatives for each function involve applying rules of differentiation systematically. Keep practicing, and soon you’ll feel like a calculus wizard!