91Ó°ÊÓ

A demand equation is an essential concept in economics. It shows the relationship between the price of a good (p) and the quantity demanded (x). In our exercise, we have been given a specific demand equation: \[ p^2 + p + 2x = 40 \] This equation tells us how the price (p) and quantity demanded (x) interact. In real-life scenarios, businesses use such equations to predict sales and set pricing strategies. They assume that by lowering the price, the demand (or quantity sold) will generally increase, and vice versa.

The chain rule is a crucial technique in calculus, especially for implicit differentiation. It helps us find the derivative of a composed function. In our demand equation, we see that the price (p) is implicitly a function of the quantity demanded (x). To apply the chain rule correctly: - Differentiate each term involving p as if p were a variable. - Multiply by the derivative of p with respect to x, i.e., \( \frac{dp}{dx} \). For instance: - Differentiating \( p^2 \): \( 2p \frac{dp}{dx} \) - Differentiating \( p \): \( \frac{dp}{dx} \) Remember also to directly differentiate any terms solely in x. In our case: - Differentiating \( 2x \) gives us 2. Combining everything, we obtain: \( 2p \frac{dp}{dx} + \frac{dp}{dx} + 2 = 0 \).

Differentiation is the process of finding the derivative of a function. In the context of our exercise, implicit differentiation allows us to determine the rate of change of one variable concerning another even when they are not separated explicitly. Let's go through the differentiation process step-by-step: 1. **Implicit Differentiation**: Take the derivative of both sides of the demand equation with respect to x. Remember that p is a function of x: \( 2p \frac{dp}{dx} + \frac{dp}{dx} + 2 = 0 \). 2. **Factor Out \( \frac{dp}{dx} \)**: Group terms involving \( \frac{dp}{dx} \) to factor it out: \( (2p + 1) \frac{dp}{dx} + 2 = 0 \). 3. **Solve for \( \frac{dp}{dx} \)**: Isolate \( \frac{dp}{dx} \) by moving other terms to the other side of the equation: \( (2p + 1) \frac{dp}{dx} = -2 \). Divide both sides by \( 2p + 1 \) to solve for \( \frac{dp}{dx} \): \( \frac{dp}{dx} = \frac{-2}{2p + 1} \). This result, \( \frac{dp}{dx} \), tells us how the price changes with respect to the quantity demanded in this particular demand equation.