1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: 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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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


    User Avatar
    Science Advisor

    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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

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


    User Avatar
    Science Advisor

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook