Math Problem Statement
Solution
The question you're working on involves finding the instantaneous rate of change of demand with respect to price for a given demand function:
where represents demand and represents price in dollars.
Part (a):
You need to find and evaluate it at .
- Derivative of with respect to :
The given function is . We need to apply the chain rule to differentiate this expression.
Taking the derivative of with respect to :
- Evaluate at :
Since , we have:
So, the instantaneous rate of change of demand when the price is dollars is .
Part (b):
You need to do the same for .
Using the derivative formula we found earlier:
Evaluate :
Since , we have:
So, the instantaneous rate of change of demand when the price is dollars is .
Interpretation:
- For part (a): The rate of change at is , meaning for every $1 increase in price, the demand drops by 2 units.
- For part (b): The rate of change at is , meaning for every $1 increase in price, the demand drops by of a unit.
Let me know if you'd like further details or explanations!
Here are some related questions:
- How do we apply the chain rule to find the derivative of functions with radicals?
- Why does the demand drop more significantly when the price is lower?
- What is the meaning of the negative sign in the rate of change?
- How would the interpretation change if the rate of change were positive?
- How can the instantaneous rate of change be used in business or economics to optimize pricing strategies?
Tip: Understanding the derivative helps analyze sensitivity—how much one variable (like demand) responds to changes in another (like price).
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Math Problem Analysis
Mathematical Concepts
Calculus
Derivatives
Instantaneous Rate of Change
Demand Function
Formulas
q(p) = 256 / sqrt(p) - 1
q'(p) = -128 / p^(3/2)
Theorems
Chain Rule of Differentiation
Suitable Grade Level
Undergraduate (Calculus I)
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