1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Difference quotient of 2^x

  1. Sep 11, 2014 #1

    mrg

    User Avatar

    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.
     
  2. jcsd
  3. Sep 11, 2014 #2
    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Difference quotient of 2^x
  1. Difference Quotients (Replies: 4)

Loading...