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Finding the derivative of an equation with an exponent

  1. Oct 13, 2007 #1
    1. The problem statement, all variables and given/known data
    Find the local max and min
    f(x)=xe^x



    2. Relevant equations
    first find the derivative of xe^x and then set it to 0



    3. The attempt at a solution
    I understand how to approach the problem but I don't know how to break apart the original equation since there is an exponent there. I know if I ln both sides I can get rid of the e and therefore bring down the x exponent, but ln(0) does not exist.
     
  2. jcsd
  3. Oct 13, 2007 #2
    to find the derivative you need to use the product rule:

    (uv)'=uv'+vu'

    and from there you can find the critical points.
     
  4. Oct 13, 2007 #3

    HallsofIvy

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    (xex)'= (x)' ex+ x(ex)'

    Now, are you saying that you do not know the derivative of ex?
     
  5. Oct 13, 2007 #4

    xe^x
    f'(x)= (e^x) (x)(e^x)
    to find the critical points:
    (e^x) (x)(e^x)=x(e^2x) = 0

    Now how do I solve for x when there is an x in the exponent?
     
  6. Oct 13, 2007 #5

    nrqed

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    You made a mistake. You must have a *sum* of two terms, not a product.
    So you must solve [itex] e^x + x e^x =0[/itex] Factor out the exponential and find when this will be equal to zero.
     
    Last edited: Oct 13, 2007
  7. Oct 13, 2007 #6
    So then I have e^x (1+x) = 0. How do I factor out the x when there is one in the exponent? I know I can take the natural log but can I do that for the 0 as well? but ln(0) does not exist.
     
    Last edited: Oct 13, 2007
  8. Oct 13, 2007 #7

    rock.freak667

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    No need to take ln...if you know that e^x is never equal to zero for all values of x, then you can simply divide by e^x
     
  9. Oct 13, 2007 #8
    You have a product with two factors; only one of them has to equal zero in order for the entire product to equal zero.
     
  10. Oct 13, 2007 #9
    Here is another way to solve for the derivitive. It is similar to the product rule, but I like it's elegance. And it has to do with the fact that [tex]\frac{d}{dx}ln(x)=\frac{1}{x}[/tex].

    [tex]f(x)=xe^x[/tex]

    Take natural log of both sides:
    [tex]ln(f(x))=ln(x)+x[/tex]

    Use the chain rule to find the derivitive of the left side, and then differentiate the right:
    [tex]\frac{f'(x)}{f(x)}=\frac{1}{x}+1[/tex]

    Since [tex]f(x)[/tex] is given above, multiply both sides by it and you end up with:
    [tex]f'(x)=e^x+xe^x[/tex]

    Now when you set the derivitive to [tex]0[/tex], you factor out the [tex]e^x[/tex] and use the zero product rule along with some function analysis to get your solution. Remember the analysis!
    [tex]0=e^x(x+1)[/tex].
     
  11. Oct 13, 2007 #10

    nrqed

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    You caNOT isolate x in that expression. So what youhave to do is to say the following: the product is zero if either of the two factors is zero. So you turn it into two separate problems. SI there any x for which [itex] e^x=0 [/itex] and is there any x for which [itex] (1+x) =0 [/itex] ?
     
  12. Oct 13, 2007 #11

    HallsofIvy

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  13. Oct 13, 2007 #12
     
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