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 the derivative \( \frac{d y}{d x} \) of each function, we'll use the product rule, chain rule, and the power rule for differentiation. Let's dive right in: First Line: For (a) \( y=4 x^{2}(5 x+3) \), applying the product rule gives us \( \frac{d y}{d x} = 4x^2 \cdot 5 + 8x(5x + 3) = 20x^2 + 40x^2 + 24x = 60x^2 + 24x \). For (b) \( y=-2 x^{3}(x+1) \), we compute \( \frac{d y}{d x} = -2x^3 \cdot 1 + (-2)(3x^2)(x + 1) = -2x^3 - 6x^2 - 2x^2 = -2x^3 - 8x^2 \). Second Line: Moving on to (c) \( y=x^{2}(1-4 x)^{4} \), we apply the product and chain rules yielding \( \frac{d y}{d x} = 2x(1-4x)^{4} + x^2 \cdot 4(1-4x)^{3}(-4) = 2x(1-4x)^{4} - 16x^2(1-4x)^{3} \). For (d) \( y=x^{2} \sqrt{1-2 x^{2}} \), we write this as \( y=x^2(1-2x^2)^{1/2} \), using both product and chain rules leads to \( \frac{dy}{dx} = 2x\sqrt{1-2x^2} + x^2 \cdot \frac{1}{2}(1-2x^2)^{-1/2}(-4x) = 2x\sqrt{1-2x^2} - \frac{2x^3}{\sqrt{1-2x^2}} \). Third Line: For (e) \( y=(4 x-3)(2 x+7)^{6} \), applying the product rule leads to \( \frac{d y}{d x} = (4)(2x + 7)^{6} + (4x - 3)(6)(2x + 7)^{5}(2) = 8(2x + 7)^{5}((4x - 3) + 12) \). Similarly, for (f) \( y=(x+5)^{3}(x-4)^{4} \), we also use product rule giving \( \frac{d y}{d x} = 3(x+5)^{2}(x-4)^{4} + 4(x+5)^{3}(x-4)^{3} \). These calculations will give you the precise rates of change for each of the functions concerning \( x \)!