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Annoying d/dx

  1. Mar 14, 2009 #1
    can you guyz help me with a quick derivative?

    eeeex
     
  2. jcsd
  3. Mar 14, 2009 #2

    gabbagabbahey

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    Gold Member

    Hi LyleX^Y, Welcome to PF!:smile:

    We're here to assist you, not to do your homework for you. As per forum rules, you must show some attempt at a solution, and are not supposed to ignore the homework help template.

    So, what have you tried?
     
  4. Mar 14, 2009 #3
    awww sorry this isnt a homework actually im just thinking of a problem my teacher would do. hehe sorry

    i got

    d/dx=(eeeex)*(d/dx eeex)*(d/dx eex)*(d/dx ex)

    i just cant get past that point i dont know how to do the product rule with 3 values.
     
  5. Mar 14, 2009 #4

    gabbagabbahey

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    Well, so far so good:smile: (although it's the chain rule you're using, not the product rule)

    Now try computing each of the derivatives in that product (use the chain rule again).... what are (d/dx eeex), (d/dx eex) and (d/dx ex)?
     
  6. Mar 14, 2009 #5
    dont i have to use the product rule for the 3 that i havent take the d/dx of?

    "(d/dx eeex), (d/dx eex) and (d/dx ex)?[/QUOTE]"
     
  7. Mar 14, 2009 #6
    could we differentiate it like a^x by so it owuld be lna*a^x
     
  8. Mar 14, 2009 #7

    Mark44

    Staff: Mentor

    No, since these aren't products -- they are function compositions, so you would need to use the chain rule. Think of it this way: if f(x) = ex, then eex = f(f(x)).
     
    Last edited: Mar 15, 2009
  9. Mar 14, 2009 #8

    Mark44

    Staff: Mentor

    I forgot to put this in my previous post. It doesn't make any sense to start and equation with d/dx. It's very much like saying [itex]\sqrt[/itex] = 3. Each of these symbols represents an operation that is yet to be performed.
     
  10. Mar 14, 2009 #9

    Mark44

    Staff: Mentor

    That's the wrong way to go, since it would make the problem more difficult for no purpose. The basic function in this problem is already "the" exponential function, ex.
     
  11. Mar 15, 2009 #10
    ok thanks..... thanks for the help... got to go do my midterm hehe thanks again
     
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