Why do we divide (-100(1.80))/(9(8000/1000) ?? Where does the 1000 come from?

College Algebra
7th Edition
ISBN:9781305115545
Author:James Stewart, Lothar Redlin, Saleem Watson
Publisher:James Stewart, Lothar Redlin, Saleem Watson
Chapter4: Exponential And Logarithmic Functions
Section: Chapter Questions
Problem 3CC: If xis large, which function grows faster, f(x)=2x or g(x)=x2?
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Why do we divide (-100(1.80))/(9(8000/1000) ?? Where does the 1000 come from?

12.4: Implici
artleby
100p + 94 ap
earch for textbooks, step-by-step explanations to homework questions
< Chapter 12.4, Proble
le...
dq
2(100p +9q
dp
Further, solve:
dq
%3D
dq
Step 2: Solve the resultant equation for
dp
dq
94
-100p
dp
-100p
99
dq
%3D
dp
Substitute 1.80 for p and 8000/1000 for q:
dq
-100(1.80)
%3D
dp
9(8000/1000)
%3D
-2.5 thousand
Here, the quantity demanded is given in thousands.
The quantity demanded is given in thousands. Therefore, this
concludes that with the increase in price by $1, the quantity
demanded decreases by approximately 2500 downloads.
Transcribed Image Text:12.4: Implici artleby 100p + 94 ap earch for textbooks, step-by-step explanations to homework questions < Chapter 12.4, Proble le... dq 2(100p +9q dp Further, solve: dq %3D dq Step 2: Solve the resultant equation for dp dq 94 -100p dp -100p 99 dq %3D dp Substitute 1.80 for p and 8000/1000 for q: dq -100(1.80) %3D dp 9(8000/1000) %3D -2.5 thousand Here, the quantity demanded is given in thousands. The quantity demanded is given in thousands. Therefore, this concludes that with the increase in price by $1, the quantity demanded decreases by approximately 2500 downloads.
< Chapter 12.4, Problem 4
The expression is 100p + 9q = 900. Also, the value of price or p is
p 3D $1.80 and that of quantity demanded in thousand or q is 8000
9
downloads.
Formula used:
Chain Rule:
If f and g are functions and
y = [g(x)i, then y = f'lg(x)] g'(x)
provided that f'Lg(x)] and g'x) exists.
Calculation:
Implicit Differentiation:
This differentiation is used to evaluate
dq
when g is defined
dp
implicitly as a function of p.
The differentiation is carried out in two steps:
Step 1: Both the sides of the given equation is differentiated with
respect to p. Since q is the function of p, always use the chain rule. In
this differentiate the equation with respect to p.
case,
(100p +94)
(900)
dp
4 (100p)+
d.
(94)
(900)
dp
dp
dp
100 (2p) +9 (2g)
da
= 0
dp
MacBook Po
%3D
Transcribed Image Text:< Chapter 12.4, Problem 4 The expression is 100p + 9q = 900. Also, the value of price or p is p 3D $1.80 and that of quantity demanded in thousand or q is 8000 9 downloads. Formula used: Chain Rule: If f and g are functions and y = [g(x)i, then y = f'lg(x)] g'(x) provided that f'Lg(x)] and g'x) exists. Calculation: Implicit Differentiation: This differentiation is used to evaluate dq when g is defined dp implicitly as a function of p. The differentiation is carried out in two steps: Step 1: Both the sides of the given equation is differentiated with respect to p. Since q is the function of p, always use the chain rule. In this differentiate the equation with respect to p. case, (100p +94) (900) dp 4 (100p)+ d. (94) (900) dp dp dp 100 (2p) +9 (2g) da = 0 dp MacBook Po %3D
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