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For each demand equation, use implicit differentiation to find \(d p / d x\). $$ (p-1)(x+5)=24 $$

Short Answer

Expert verified
\( \frac{dp}{dx} = \frac{-(p-1)}{x+5} \).

Step by step solution

01

- Differentiate both sides with respect to x

To find the derivative \( \frac{dp}{dx} \), we need to implicitly differentiate the given equation. Let's differentiate both sides of the equation: \((p-1)(x+5) = 24\)Differentiating the left side using the product rule, where if \( y = uv \), then \( \frac{dy}{dx} = u'v + uv' \), we get:\[ \frac{d}{dx}((p-1)(x+5)) = (p-1)'(x+5) + (p-1)(x+5)' \] Note: The derivative of a constant (24) is 0.
02

- Apply the Derivative Rules

The derivative of \((p-1)\) with respect to \(x\) is \( dp/dx \), as it contains \( p \), so \((p-1)' = dp/dx \). The derivative of \((x+5)\) with respect to \(x\) is 1, as it's linear. So, the product rule gives:\[ \frac{d}{dx}((p-1)(x+5)) = \frac{dp}{dx}(x+5) + (p-1) \cdot 1 \]
03

- Set the Derivatives Equal to Zero

Since the differentiation of the right side, which is 24, results in 0, set the differentiated left side equal to 0:\[ \frac{dp}{dx}(x+5) + (p-1) = 0 \]
04

- Solve for dp/dx

Rearrange the equation to isolate \( \frac{dp}{dx} \):\[ \frac{dp}{dx}(x+5) = -(p-1) \] Now, divide both sides by \((x+5)\) to solve for \( \frac{dp}{dx} \):\[ \frac{dp}{dx} = \frac{-(p-1)}{x+5} \]
05

- Simplify the Final Expression

The expression obtained is the implicit derivative \( \frac{dp}{dx} \). Simplify or rearrange it if necessary for further interpretation:\[ \frac{dp}{dx} = \frac{-(p-1)}{x+5} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Product Rule
The product rule is a fundamental principle in calculus used to differentiate functions that are the product of two or more expressions. Imagine you have two functions, such as \( u(x) = (p-1) \) and \( v(x) = (x+5) \). When these functions are multiplied, their derivative is not simply the product of their individual derivatives. Instead, the product rule provides a way to find the correct derivative when two functions are multiplied.

In mathematical terms, if \( y = u \cdot v \), then the derivative \( \frac{dy}{dx} \) is calculated as:
  • \( u'v + uv' \)
This means you take the derivative of the first function \( u \), multiply it by the second function \( v \), and add the product of the first function \( u \) and the derivative of the second function \( v \).

When applying the product rule to the demand equation \((p-1)(x+5)\), you end up differentiating each term appropriately. The process helps in finding how a change in one variable affects another in complex products.
Derivative Rules
Understanding derivative rules helps to make differentiation straightforward. There are several basic rules of differentiation that you consistently use.
  • **Constant Rule:** The derivative of a constant is zero. If \( c \) is a constant, then \( \frac{d}{dx}[c] = 0 \).
  • **Power Rule:** For any real number \( n \), \( \frac{d}{dx}[x^n] = nx^{n-1} \).
  • **Sum Rule:** The derivative of a sum equals the sum of derivatives, \( \frac{d}{dx}[u + v] = \frac{du}{dx} + \frac{dv}{dx} \).
  • **Product Rule:** As previously explained, use when differentiating a product of terms.
  • **Chain Rule:** Useful for differentiating composite functions, \( \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \).
In the context of implicit differentiation, these rules help in tackling equations where one variable is a function of another. For example, in \( (p-1)(x+5) = 24 \), you're not given \( p \) as an explicit function of \( x \), so implicit differentiation and these rules let you find \( \frac{dp}{dx} \) more effectively.
Demand Equation
In economics, a demand equation outlines the relationship between the quantity demanded of a good and its price. It considers various factors such as population segment, income level, or price changes.

The given demand equation \((p-1)(x+5) = 24\) is an implicit form that relates price \( p \) and quantity \( x \). Solving for \( \frac{dp}{dx} \), using implicit differentiation, helps to understand how demand \( x \) changes with respect to price \( p \). Applying implicit differentiation provides insights into elasticity and slope of the demand curve, which are vital in predicting consumer behavior.

In real-world applications, such equations help firms set prices or assess potential revenue from changes in market conditions. They provide a basis for further economic analysis by understanding the gradient of change between variables, like price and demand.

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