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Need to find the second derivative of:

  1. Aug 4, 2010 #1
    f(x)= 1/125(e5x)(5x-2)


    I think this is the first derivative but I ain't good at math and gives me some headaches I used the product rule.. but still I have doubts =(

    f '(x)= 1/125(5)(e5x)(5x-2) + (5)(1/125(e5x)

    Please help me!! :cry:
    I need to understand how to do this
     
    Last edited: Aug 4, 2010
  2. jcsd
  3. Aug 4, 2010 #2
    Assuming you have [tex] f(x) = \frac{1}{125}e^{5x}(5x - 2), [/tex] you need to apply the product rule.

    Given [tex] f(x) = g(x)h(x), [/tex]

    [tex] f'(x) = g'(x)h(x) + g(x)h'(x). [/tex]

    And you'll need the product rule again to find f''(x). Also, I'd leave the constant 1/125 out in front while taking the derivatives until the end.
     
  4. Aug 4, 2010 #3
    Thanks Raskolnikov! Now I only need to identify which rule I need to use in each case.
     
  5. Aug 4, 2010 #4
    Yep, you got it right (though I think you're overloading on parentheses :P)! For the second derivative, you have to use the product rule again on the first term since it has both a factor of e^5x and a factor of (5x - 2). However, you don't need the product rule for the 2nd term since it's just e^5x.

    Try to find f''(x). Just post whatever answer you get or where you're getting stuck, and I'll help you from there if you need.
     
  6. Aug 4, 2010 #5
    I'm not sure about this... as I said.. I haven't mastered this rules :cry: and is getting really messed up

    f"(x)= ([tex]\frac{1}{125}[/tex])(25)e[tex]^{5x}[/tex](5x-2) + 5([tex]\frac{1}{125}[/tex](5)e[tex]^{5x}[/tex]) + ([tex]\frac{1}{125}[/tex](5)e[tex]^{5x}[/tex])....

    I dont know... think I got confused :confused:
     
  7. Aug 4, 2010 #6
    Close! You forgot another factor of 5 on the last term. Don't worry, you'll get better with practice.

    So it should simplify to:

    [tex] f''(x) = \frac{1}{5}e^{5x}(5x - 2) + \frac{2}{5}e^{5x}. [/tex]
     
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