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Product rule calculus help

  1. Mar 5, 2005 #1
    y = x ln x

    [tex]\frac{dy}{dx} = \frac{1}{x ln x} \cdot 1[/tex]

    Is that correct?
     
  2. jcsd
  3. Mar 5, 2005 #2
    wrong... use product rule

    [tex] \frac{d}{dx} (fg) = \frac{df}{dx}g+f\frac{dg}{dx} [/tex]
     
  4. Mar 5, 2005 #3
    And how would that look like? Which is f & which one is g? :confused:
     
  5. Mar 5, 2005 #4
    Hmm i'm not great at this but,

    y = x lnx
    I was gunna say.. that whenever you take the derivative of logs.. 1/function * derivative of that function
    so that'd give you
    y = (1/xlnx)(1/x)
    but by using the product rule...
    y = 1(xlnx) + x(1/x))
    Hmm, definately don't listen to me, i don't know WHats going on. Here to learn!
     
  6. Mar 5, 2005 #5
    I was taught that way. How did you get 1/x?
     
  7. Mar 5, 2005 #6
    hmm I think i might just be being stupid. 1/x is the known derivative of lnx.. but i guess with that other x there.. x lnx, you would need to use product rule like that guy said?
     
  8. Mar 5, 2005 #7
    Oh I see...
    How do you know it?
     
  9. Mar 5, 2005 #8
    Use product rule. f(x)=x and g(x)=ln(x)
    Now just find derivatives
    y = x'*lnx + x*ln(x)'
     
  10. Mar 5, 2005 #9
    Yeah I got it. Thank you.
     
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