Math Problem Statement
7,8,9
Solution
Let's break down and solve the problems from the image.
Question 7:
(a) The demand function , find .
This is a basic differentiation problem. Differentiating with respect to :
(b) The cost function , find .
Differentiating with respect to :
Question 8:
(a) Show that .
This expression can be simplified using the difference of squares:
(b) If , show that .
We need to compute . Multiply the numerator and denominator by :
(c) Use the result in part (b) to show that for , implies .
The derivative of is the limit as of the expression in part (b):
(d) Show that the result could also be written as .
We know that . Using power rule for differentiation:
Question 9:
(a) If , show that
First, expand :
Now subtract :
Now divide by :
(b) Hence, .
As , the terms involving vanish, and we are left with:
Would you like further details on any part of these solutions?
Related Questions:
- How do you find the second derivative of ?
- What is the interpretation of a demand function in economics?
- How do you apply the product rule to differentiate a product of two functions?
- Can you find the critical points of ?
- What is the difference between marginal cost and marginal revenue in economic terms?
Tip:
Always check the domain of the function when differentiating, especially with square roots and logarithms, to avoid undefined regions.
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Math Problem Analysis
Mathematical Concepts
Differentiation
Demand and cost functions
Algebraic manipulation (difference of squares)
Power rule
Formulas
Demand function: D(P) = a - bP
Cost function: C(x) = p + qx^2
Derivative of a square root function: d(√x)/dx = 1 / (2√x)
Derivative of a power: d(x^n)/dx = nx^(n-1)
Difference of squares: (A - B)(A + B) = A^2 - B^2
Theorems
Difference of squares
Power rule for differentiation
Suitable Grade Level
Grades 10-12
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