91Ó°ÊÓ

A demand equation is an essential part of understanding economics, as it helps represent the relationship between the quantity demanded (often denoted as \( x \)) and the price (often denoted as \( p \)). This relationship can be complex and non-linear, as seen in the given equation \[ p^3 + p - 3x = 50 \]where the variables \( p \) and \( x \) are not easily isolated. Without an explicit solution for \( p \), we must use implicit differentiation to explore how changes in \( x \) affect the price \( p \). Implicit differentiation can unravel the interconnected nature of these variables and provide actionable insights, such as marginal changes in price as demand shifts. Understanding this relationship through a demand equation is fundamental for making informed economic decisions.

The chain rule is a pivotal calculus tool needed for differentiating composite functions, especially useful in implicit differentiation. In the exercise, \( p \) is treated as a function of \( x \) when differentiating. The chain rule states that:

• If \( u = f(v) \) and \( v = g(x) \), then the derivative of \( u \) with respect to \( x \) is:
• \( \frac{du}{dx} = \frac{du}{dv} \times \frac{dv}{dx} \)

When applying this to the term \( p^3 \), it becomes:

• \( 3p^2 \frac{dp}{dx} \)

Where \( \frac{dp}{dx} \) represents the derivative of \( p \) concerning \( x \). This application shows how small changes in \( x \) translate into changes in \( p \) through composite structures. Thus, the chain rule facilitates understanding the dependencies in implicitly defined relationships.

Differentiating the demand equation involves methodically applying derivative rules to each term. Here’s a step-by-step breakdown. For the left side:

• \( p^3 \) becomes \( 3p^2 \frac{dp}{dx} \)
• \( p \) becomes \( \frac{dp}{dx} \)
• \(-3x \) becomes \(-3 \)

For the right side:

• The constant 50 differentiates to 0.

Combining these gives the equation:\[ 3p^2 \frac{dp}{dx} + \frac{dp}{dx} - 3 = 0 \]Next, factor \( \frac{dp}{dx} \) out:\[ \frac{dp}{dx}(3p^2 + 1) = 3 \]Finally, solve for \( \frac{dp}{dx} \):\[ \frac{dp}{dx} = \frac{3}{3p^2 + 1} \]This process reveals \( \frac{dp}{dx} \), illustrating how changes in demand (\( x \)) affect the price (\( p \)), thereby offering vital indicators for economic analysis.