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Derivative of e^x

  1. Feb 9, 2005 #1

    Aki

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    I don't understand why the derivative of e^x is e^x itself.
     
  2. jcsd
  3. Feb 9, 2005 #2
    Have you tried apply the definition of a derivative for [itex]f(x) = e^x[/itex]?

    [tex]\frac{d}{dx}f(x) = \lim_{h\to 0}\frac{f(x+h) - f(x)}{h}[/tex]

    --J
     
  4. Feb 9, 2005 #3

    matt grime

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    There are two types of answer to this.

    1. Because it's the definition: exp is the function satisfying th differential equation f' = f. The question then becomes: why does it have this form, ie an exponential. If you were to work out the slope using Justin's hint then you'd see that, assuming some limits exist, all exponential functions have derivatives proportional to themselves, and e is the number where the proportionality constant is 1.

    2. Assume the definition of exp{x} is its power series, abuse analysis by differentiating term by term.
     
  5. Feb 10, 2005 #4
    first consider the following

    the integral of In(x) is equal to xInx-x.
    the value of the graph of e^x is the same as doing the e^x bound on the Inx graph. so we could plug in e^x for x into xInx-x. Giving xe^x-e^x as the area inside of the Inx graph. Now this is below the curve, we are trying to find the area above the curve on the Inx graph. So the rectangle area is equal to x(e^x). And the area of the region is now xe^x-(xe^x-e^x)=e^x

    I hope this logic was right.
     
  6. Feb 10, 2005 #5

    dextercioby

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    Not really.I failed to see the connection between the area under one of the 2 curves and the derivative of e^{x}.

    Daniel.
     
  7. Feb 10, 2005 #6

    mathwonk

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    perhaps we should ask what the questioner knows about the function e^x.

    i.e. the respinders are observing that if you know two things:

    1) the definition of a derivative

    2) the property e^(x+h) = e^x e^h,

    then the result is essentially forced on you.
     
  8. Feb 10, 2005 #7
    my post wasnt very clear, but the region of area above the Inx graph resembles the area under the curve in the e^x graph. it creates a rectangle.
     
    Last edited: Feb 10, 2005
  9. Feb 10, 2005 #8

    dextercioby

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    Maybe mine wasn't very clear.What does that fact (resemblence of areas beneath the graphs) gotta do with
    [tex] (e^{x})'=e^{x} [/tex] ?

    Daniel.
     
  10. Feb 10, 2005 #9
    the antiderivative of e^x is equal to e^x+C. The derivative of the antiderivative is equal to e^x.
     
  11. Feb 10, 2005 #10
    I have a related question.

    How can you show that the equation f'(x) = f(x) has a unique solution?
     
  12. Feb 10, 2005 #11

    dextercioby

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    Gues what,it DOES NOT...Ups,i didn't notice Matt forgot about the initial condition (which would select e^{x} from the infinity of solutions of the ODE) f(0)=1 [/tex]

    Daniel.
     
  13. Feb 10, 2005 #12
    first i thought he was asking why the integral of e^x was e^x? Then i reread it and saw that he was actually asking why e^x=(e^x)'
    After this i did the pathetic "antiderivative of a function is equal to the integral, which is equal to the area under the curve business."
    But i thought of the method of similarities in inverse curves, finding the integral, which i thought was pretty. I was asked to find the area under the curve in the arctan on the ap test. so i had make something.
     
  14. Feb 10, 2005 #13
    i proved by the Inx graph, that the integral of e^x is e^x+C.
     
  15. Feb 10, 2005 #14

    dextercioby

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    However,there's a long way between the derivative of e^{x} and the area under the graph of a function...Almost infinitely long...

    Daniel.
     
  16. Feb 10, 2005 #15
    one can only confuse derivatives and integrals only when dealing with e^x. I shouldnt drop the C, which is the constant. I have a question, that probably most of you guys with think as easy, but is there any other function where the derivative of the function is equal to the function itself? prove or disprove.
     
    Last edited: Feb 10, 2005
  17. Feb 10, 2005 #16

    dextercioby

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    No.e^{x} is the only one.Matt suggested via the ODE.And i added the initial condition [itex] y(0)=1 [/itex].

    Daniel.
     
  18. Feb 10, 2005 #17

    mathwonk

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    assume f' = f, and take the derivative of f/e^x. se what happens. then use the MVT.
     
  19. Feb 11, 2005 #18
    how about -e^(x)


    or d/dx (y.e^(x)) where y is a arbitrary constant
     
  20. Feb 11, 2005 #19

    matt grime

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    Differential equations have the full complement of solutions that are uniquely determined according to whether they satsify the Lipschitz Condition and something else. I'm sure Wolfram will provide the answer as ever.
     
  21. Feb 11, 2005 #20
    This is not an easy task, but in analysis it essentially involves making precise the intuitive argument that "if we take small steps and adjust our course according to the DE then the path we walk will be uniquely determined by where we start."

    I believe the uniqueness theorem for linear ODEs can be proven through linear algebra, the set of solutions form a vector space yadda yadda.
     
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