How can the difference quotient be used to simplify the expression for y=2^x?

Click For Summary
SUMMARY

The discussion focuses on simplifying the difference quotient for the function y=2^x using the formula (f(x+h)-f(x))/h without employing the chain rule or other advanced techniques. Participants highlight the challenge of eliminating the variable h from the denominator, ultimately leading to the expression $$\frac{2^x\ln 2(e^{h\ln 2}-1)}{h\ln 2}$$. The conversation emphasizes that while limits can be approached using known results, such as $$\lim_{u\rightarrow 0}\frac{e^{u}-1}{u}=1$$, a straightforward algebraic simplification to directly substitute h=0 is not feasible. This problem serves as a valuable exercise for students to enhance their understanding of difference quotients.

PREREQUISITES
  • Understanding of exponential functions, specifically y=2^x
  • Familiarity with the difference quotient formula (f(x+h)-f(x))/h
  • Basic knowledge of limits, particularly $$\lim_{u\rightarrow 0}\frac{e^{u}-1}{u}$$
  • Awareness of logarithmic properties and their applications
NEXT STEPS
  • Study the properties of exponential functions and their derivatives
  • Learn about the application of limits in calculus, focusing on $$\lim_{h\rightarrow 0}\frac{a^{h}-1}{h}$$
  • Explore advanced techniques for simplifying difference quotients in calculus
  • Investigate the relationship between logarithmic and exponential functions in depth
USEFUL FOR

Students learning calculus, educators seeking to challenge students with difference quotient problems, and anyone interested in the foundational concepts of limits and derivatives in exponential functions.

mrg
Messages
16
Reaction score
0
1. Simplify the algebraic expression you get for Δy and Δy/Δx for the equation y=2^x



2. Use the difference quotient (f(x+h)-f(x))/h. No use of chain rule or other shortcuts.



3. I've tried a host of things, including raising terms to a natural log power (I.e. e^(ln2)*x*h), using logarithm properties, simplifying things... It seems that I cannot get rid of an h in some denominator. Now, this is a problem well before we actually learn derivatives, so things like that chain rule and implicit differentiation haven't been learned. This is a problem to challenge the kids with their difference quotient skills. I'm beginning to fear that it can't be done.
 
Physics news on Phys.org
There is no way to get rid of the ##h## in the denominator. The best you can do is either write it as $$\frac{2^x\ln 2(e^{h\ln 2}-1)}{h\ln 2}=\frac{2^x\ln 2(e^{u}-1)}{u}$$ and go from there, knowing that ##\lim_{u\rightarrow 0}\frac{e^{u}-1}{u}=1## or, if you know that $$\lim_{h\rightarrow 0}\frac{a^{h}-1}{h}=\ln a$$ (i.e. the general version of the known limit) then you can use that.

But there is no way to algebraically simplify this limit so that you can just plug in ##h=0## like you can with the derivative limits for the algebraic functions.

FYI, you need some "advanced tech" for the trig derivatives as well.
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
Replies
5
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 24 ·
Replies
24
Views
3K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K