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Derivative of 2^x

  1. Feb 6, 2012 #1
    1. The problem statement, all variables and given/known data

    Let's take two problems, the derivative of




    For the second one the book says that answer is


    Well if you can do that for the above, then why not for the first problem?

    The book gives the answer to the first problem as

    2x ln 2

    why not x2x-1?

    essentially what i need to know for finding derivatives, there are two techniques.
    1. nx^n-1
    2. k^x ln k

    when do I use which technique?
  2. jcsd
  3. Feb 6, 2012 #2


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    Homework Helper

    Because that rule only works when the power of the x term is a constant (independent of x).

    To differentiate 2x, express it as exln(2). This is of the form ekx, where k is a constant.
  4. Feb 6, 2012 #3


    User Avatar
    Science Advisor

    it makes a BIG difference whether x is "upstairs" (in the exponent), or "downstairs" (being exponentiated).

    if you go back to the definition:

    d(2x)/dx = limh→0 (2(x+h)-2x)/h

    you can see that you're not going to get an easy way to simplify.

    basically, e is the "natural base" for exponential functions, and other bases have logarithms as a "conversion factor":

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

    which is of the form eax, so has derivative aeax, by the chain rule.
  5. Feb 6, 2012 #4


    Staff: Mentor

    2x and xn are very different types of functions. The first is an exponential function, so called because the variable is in the exponent. The second is a power function, so called because the variable is in the base, which is raised to a fixed power.

    There is no single differentiation rule, other than the definition, that covers both of these functions. You are misusing the power rule to conclude that d/dx(2x) = x2x-1. Each of the rules has "fine print" that says when it can be applied. Read the fine print.
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