Find dy/dx: 2x + 3xy = 2y2

We will apply the differentiation rules:

• $\frac{d}{dx}(c) = 0$
• $\frac{d}{dx}(cx) = c$
• $\frac{d}{dx}(cy) = c\frac{dy}{dx}$
• $\frac{d}{dx}(u \cdot v) = u'v + uv'$ (product rule)
• $\frac{d}{dx}(y^n) = ny^{n-1}\frac{dy}{dx}$ (chain rule)

Differentiating each term:

• For $2x$: $\frac{d}{dx}(2x) = 2$
• For $3xy$: Use the product rule with $u=3x$ and $v=y$. $u' = \frac{d}{dx}(3x) = 3$ $v' = \frac{d}{dx}(y) = \frac{dy}{dx}$ So, $\frac{d}{dx}(3xy) = (3)(y) + (3x)\left(\frac{dy}{dx}\right) = 3y + 3x\frac{dy}{dx}$
• For $2y^2$: Use the chain rule. $\frac{d}{dx}(2y^2) = 2 \cdot 2y^{2-1} \cdot \frac{dy}{dx} = 4y\frac{dy}{dx}$

Subtract $3x\frac{dy}{dx}$ from both sides:

$$2 + 3y = 4y\frac{dy}{dx} - 3x\frac{dy}{dx}$$

$$\frac{dy}{dx} = \frac{2 + 3y}{4y - 3x}$$

The derivative $\frac{dy}{dx}$ is:

$$\frac{2 + 3y}{4y - 3x}$$