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Homework Help: The derivative of 2^x

  1. May 24, 2012 #1
    1. The problem statement, all variables and given/known data

    I'm trying to find the derivative of 2^x by hand so that I can better understand the number e. This video

    says the answer is .69. I can't figure out how to get that

    2. Relevant equations

    lim h -> 0 [f(x0+h) - f(x0)]/h

    3. The attempt at a solution

    So let's take point (3,8)

    [2(3+h)^3 - 4(3)^3]/h

    = [54 + 2h^3 - 108]/h

    = 54

    not exactly .69
  2. jcsd
  3. May 24, 2012 #2
    I am not sure what you are doing here - but that looks incorrect.

    The easiest way to do this derivative is to write y = 2^x and take the natural log of both sides and then differentiate.
  4. May 24, 2012 #3
    If I do that, I get

    ln y = x ln 2

    Two unknowns and one equation, can't be solved.
  5. May 24, 2012 #4
    No... you take the derivative with respect to x of both sides.
  6. May 24, 2012 #5
    Then I will still have an x in my solution, not .69
  7. May 24, 2012 #6
    wait, the natural log of 2 is .69, ok, I get it now.
  8. May 24, 2012 #7
    Are you sure? Because the derivative of 2^x is not .69. It's (ln2)*2^x. It is only .69 at x=0.
  9. May 24, 2012 #8
    well, i get enough to satisfy me for the moment.
  10. May 24, 2012 #9
    well, ok then.
  11. May 24, 2012 #10
    This is wrong on so many levels... You should really read through the more basic stuff again...

    So if f(x) = 2x, then f(x0+h)-f(x0) = 2(x0+h)-2x0

    Next, you don't pick a random value for h... Instead, you take the limit as h tends to zero of (f(x+h)-f(x))/h
  12. May 24, 2012 #11


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    Gold Member

    Or you could just learn the general formula: ##\frac{d(a^x)}{dx}=a^x.\ln a## where a is any real constant.
  13. May 24, 2012 #12


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    Science Advisor

    Your whole question doesn't make much sense. The derivative of [itex]2^x[/itex] is a function of x, not a number. Since, in fact, the derivative of [itex]2^x[/itex] is the function [itex]ln(2)2^x[/itex], and ln(2) is (approximately) 0.69, the derivative of [itex]2^x[/itex] at x= 0 is (approximately) 0.69.

    Pretty much everything here is wrong. For one thing, [itex](x+ h)^3= x^3+ 3x^2h+ 3xh^3+ h^3[/itex], NOT "[itex]x^3+ h^3[/itex]". But where did that [itex]2(3+h)^3[/itex] come from anyway? The difference quotient for [itex]2^x[/itex] would be
    [tex]\frac{2^{x+ h}- 2^x}{h}[/tex]
    not what you have. (Which looks like it would be for [itex]2x^3[/itex] if that "4" were a "2".)

    We could then write [itex]2^{x+y}= 2^x2^h[/itex] and factor [itex]2^x[/itex] out. The difference quotient becomes
    [tex]2^x \frac{2^h- 1}{h}[/tex]
    so that the derivative, the limit, as h goes to 0, of that, is [itex]2^x[/itex] times the limit of that last fraction. In fact, it is easy to see that the derivative of [itex]a^x[/itex] is just [itex]a^x[itex] itself times the limit of
    [tex]\frac{a^h- 1}{h}[/tex]

    "e" happens to have that limit equal to 1. And one can use the properties of [itex]e^x[itex] to show that
    [tex]\lim_{h\to 0}\frac{a^h- 1}{h}= ln(a)[/tex]

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