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Derivative of 2^x

  1. Jan 5, 2010 #1
    Hi, i've been trying to find the derivative of 2^x and i got stuck here:
    Lim (2^x - 1)/x

    How can i solve this limit (with steps please)
    ps. im only 17
  2. jcsd
  3. Jan 5, 2010 #2


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    Have you had the same limit for ex? What you want to do is notice that

    2x = (eln(2))x = exln(2)

    Now your limit becomes:

    [tex]\lim_{x\rightarrow 0}\frac{e^{x\ln 2} - 1}{x}[/tex]

    Now let u = x ln(2) and use what you know about ex.

    [Edit] Corrected typo
    Last edited: Jan 5, 2010
  4. Jan 5, 2010 #3
    I have been doing it this way all the time which I thought it is very straight forward:

    [tex]y=2^{x}\Rightarrow ln(y)=xln(2)[/tex]

    [tex]\Rightarrow \frac{1}{y}\frac{dy}{dx}=ln(2)[/tex]

    [tex]\Rightarrow dy=d[2^{x}]=yln2dx=2^x ln(2)dx[/tex]
    Last edited: Jan 5, 2010
  5. Jan 5, 2010 #4


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    No, but probably a typo. You have to multiply both sides by y dx, not x dx.
    [itex]dy= yln(2)dx= 2^x ln(2) dx[/itex].
  6. Jan 5, 2010 #5
    With [tex]e^{ln(2)x}[/tex], you can easily take the derivative of that. Just don't simplify the [tex]e^{ln2}[/tex] part. You should get [tex]ln(2)e^{ln(2)x}[/tex] as the numerator while the denominator becomes 1. Now do this expression with [tex]\lim_{x->0}[/tex]. By the way, LCKurtz made a slight typo.
  7. Jan 5, 2010 #6
    My finger is get too far ahead of me!!! And I have destroy evidence already!!!:wink:
  8. Jan 7, 2010 #7
    I guess the answer is ln(2) * 2^x
    No wonder i couldnt solve it before, they thaught us nothing about e

    Depending on what you've said this means the derivative of e^x is e^x.. right?
  9. Jan 7, 2010 #8


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    Yes. Most calculus books do that first before trying other bases.
  10. Jan 8, 2010 #9


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    This problem might have been given in preparation for introducing "e".

    The derivative of [itex]2^x[/itex] is
    [tex]\lim_{h\to 0}\frac{2^{x+y}- 2^x}{h}= \lim_{h\to 0}\frac{2^x2^h- 2^x}{h}[/tex]
    [tex]= \lim{h\to 0}\left(\frac{2^h- 1}{h}\right)2^x[/itex]
    [tex]= \left(\lim_{h\to 0}\frac{2^h- 1}{h}\right)2^x[/tex]
    That is, of course, a constant times [itex]2^x[/itex].

    Similarly, the derivative of [itex]3^x[/itex] is
    [tex]= \left(\lim_{h\to 0}\frac{3^h- 1}{h}\right)3^x[/tex]
    a constant times [itex]3^x[/itex]

    In that same way you can show that the derivative of [itex]a^x[/itex], for a any positive real number, is [itex]C_a a^x[/itex].
    Further, by numerical approximations, you can show that [itex]C_2[/itex] is less than 1 and [itex]C_3[/itex] is greater than 1. There exists, then, a number, a, between 2 and 3 such that [itex]C_a= 1[/itex]. If we call that number "e", then the derivative of [itex]e^x[/itex] is just [itex]e^x[/itex] itself.
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