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

To find the derivatives of the given functions, we can apply common rules of differentiation like the chain rule, quotient rule, and product rule. Let's get ready to dive into some calculus magic! 1. For \( y = (5 - 2x^2)^6 \), we use the chain rule: \[ y' = 6(5 - 2x^2)^5 \cdot (-4x) = -24x(5 - 2x^2)^5 \] 2. For \( y = \frac{1}{(\sqrt{x} - x)^3} \), we will apply the quotient rule: \[ y' = -\frac{3(\sqrt{x} - x)^{-4}}{2\sqrt{x} - 1} \cdot \left(\frac{1}{2\sqrt{x}} - 1\right) \] 3. For \( y = \frac{3}{2\sqrt{x+1}} \), we can rewrite it as \( y = 3(2(x+1)^{1/2})^{-1} \) and apply the chain rule: \[ y' = -\frac{3}{2} \cdot (2(x + 1)^{-\frac{1}{2}}) \cdot \frac{1}{2} = -\frac{3}{4\sqrt{x + 1}} \] 4. For \( y = (4x - 3)(1 - 2x)^4 \), we apply the product rule: \[ y' = (4)(1 - 2x)^4 + (4x - 3)(4(1 - 2x)^3 \cdot (-2)) = 4(1 - 2x)^4 - 8(4x - 3)(1 - 2x)^3 \] 5. For \( y = \frac{2x - 3}{x^2 - 5} \), we will apply the quotient rule: \[ y' = \frac{(2)(x^2 - 5) - (2x - 3)(2x)}{(x^2 - 5)^2} = \frac{2x^2 - 10 - 4x^2 + 6x}{(x^2 - 5)^2} = \frac{-2x^2 + 6x - 10}{(x^2 - 5)^2} \] 6. For \( y = (x^3 + 1)\sqrt{2x + 7} \), we use the product rule: \[ y' = (3x^2)(\sqrt{2x + 7}) + (x^3 + 1)(\frac{1}{2\sqrt{2x + 7}} \cdot 2) = 3x^2\sqrt{2x + 7} + \frac{(x^3 + 1)}{\sqrt{2x + 7}} \] And there you have it! The derivatives of all the given functions, fresh off the calculus press. Happy differentiating!