Differentiate the following functions using the chain rule: (a) U = (2x + 3)2 (b) V = (3x − 1)^(2/3) (c) y = 1/(3x+2) (d) y = 1/(3x+2)2

Let $u = 2x+3$.

$$\frac{dU}{dx} = 2(2x+3)^{2-1} \cdot \frac{d}{dx}(2x+3)$$

$$\frac{dU}{dx} = 2(2x+3) \cdot (2) = 4(2x+3) = 8x + 12$$

The derivative is $\boxed{\mathbf{8x + 12}}$.

b) $V = (3x - 1)^{2/3}$

Let $u = 3x-1$.

$$\frac{dV}{dx} = \frac{2}{3}(3x-1)^{\frac{2}{3}-1} \cdot \frac{d}{dx}(3x-1)$$

$$\frac{dV}{dx} = \frac{2}{(3x-1)^{1/3}}$$

The derivative is $\boxed{\mathbf{\frac{2}{(3x-1)^{1/3}}}}$.

c) $y = \frac{1}{3x+2}$

Let $u = 3x+2$.

$$\frac{dy}{dx} = -1(3x+2)^{-1-1} \cdot \frac{d}{dx}(3x+2)$$

$$\frac{dy}{dx} = -\frac{3}{(3x+2)^2}$$

The derivative is $\boxed{\mathbf{-\frac{3}{(3x+2)^2}}}$.

d) $y = \frac{1}{(3x+2)^2}$

Let $u = 3x+2$.

$$\frac{dy}{dx} = -2(3x+2)^{-2-1} \cdot \frac{d}{dx}(3x+2)$$

$$\frac{dy}{dx} = -\frac{6}{(3x+2)^3}$$

The derivative is $\boxed{\mathbf{-\frac{6}{(3x+2)^3}}}$.

e) $y = \frac{1}{\sqrt{3x^2+2x-1}}$

Let $u = 3x^2+2x-1$.

$$\frac{dy}{dx} = -\frac{1}{2}(3x^2+2x-1)^{-\frac{1}{2}-1} \cdot \frac{d}{dx}(3x^2+2x-1)$$

$$\frac{dy}{dx} = -\frac{3x+1}{(3x^2+2x-1)^{3/2}}$$

The derivative is $\boxed{\mathbf{-\frac{3x+1}{(3x^2+2x-1)^{3/2}}}}$.

f) $y = \sqrt{3x^2+2x-1}$

Let $u = 3x^2+2x-1$.

$$\frac{dy}{dx} = \frac{1}{2}(3x^2+2x-1)^{\frac{1}{2}-1} \cdot \frac{d}{dx}(3x^2+2x-1)$$

$$\frac{dy}{dx} = \frac{3x+1}{\sqrt{3x^2+2x-1}}$$

The derivative is $\boxed{\mathbf{\frac{3x+1}{\sqrt{3x^2+2x-1}}}}$.

g) $y = (3\sqrt{x} - 2x)^3$

Let $u = 3x^{1/2} - 2x$.

$$\frac{dy}{dx} = 3(3x^{1/2} - 2x)^{3-1} \cdot \frac{d}{dx}(3x^{1/2} - 2x)$$

$$\frac{dy}{dx} = 3(3x^{1/2} - 2x)^2 \left(\frac{3}{2}x^{-1/2} - 2\right)$$ $$\frac{dy}{dx} = 3(3\sqrt{x} - 2x)^2 \left(\frac{3}{2\sqrt{x}} - 2\right)$$

The derivative is $\boxed{\mathbf{3(3\sqrt{x} - 2x)^2 \left(\frac{3}{2\sqrt{x}} - 2\right)}}$.