   Chapter 4.5, Problem 74E ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

#### Solutions

Chapter
Section ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919
Textbook Problem
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# Demand In Exercises 73 and 74, find dx/dp for the demand function. Interpret this rate of change for a price of $10. x = 500 ln ( p 2 + 1 ) To determine To calculate: The value of dxdp for the demand function which is approximated by x=500ln(p2+1). And explain the same for the change of a price of$10 if the demand function is as follows: Explanation

Given information:

The provided function is x=500ln(p2+1) and graph is,

Formula used:

The derivative of function f(x)=un using the chain rule is:

f(x)=ddx(un)=nun1dudx

Where, u is the function of x.

Let u be a differentiable function of x then ddx[lnx]=1x,x>0 and ddx(lnu)=1ududx,u>0

As with the natural logarithm function if x and y is a positive number such that x1 and y1 then logxy=lnxlny

Calculation:

Consider the function x=500ln(p2+1),

Apply exponential properties to the function x=500ln(p2+1),

x=500ln(p2+1)x=500(ln(p2+1))1

The derivative of function x=500ln(p2+1) is calculated as follows:

Apply chain rule of derivative to the function,

dxdp=ddp500(ln(p2+1))<

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