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Difference quotient of 2^x

  1. Sep 11, 2014 #1


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    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.
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