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Derivatives of natural numbers

  1. Nov 6, 2008 #1
    1.
    I was trying to understand the proof of
    (d/dx) b^x = (ln b)*(b^x)
    it says:
    b= e^(ln b)
    so, b^x= e^((ln b)*x))
    So now we use the chain rule:
    (d/dx) b^x = (d/dx) e^((ln b)*x))

    I understand everything so far, but not the next step.
    It says then that
    (d/dx) e^((ln b)*x))= (ln b)*e^(ln b)*x)

    how did they get this, i am confused about the bold part. I am forgetting something about taking the derivative of the natural number 'e'?
    I know that d/dx e^x = x

    so how does the ln b come behind e^x above?

    what does (d/dx) e^f(x) equal to?

    is (d/dx) e^f(x) = f '(x)* e^f(x) ?
    if so then
    1 + ln b
    should come behind that because
    d/dx (ln b)*x is equal to 1 + ln b


    Please help me
    thank you
     
  2. jcsd
  3. Nov 6, 2008 #2
    uhh

    hello?
     
  4. Nov 6, 2008 #3
    ok i found out that
    (d/dx) e^f(x) = f '(x)* e^f(x)

    still i am confused about the other thing.
     
  5. Nov 6, 2008 #4

    Dick

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    d/dx(ln(b)*x)=ln(b). Not ln(b)+1. Why would you think it's that? b is a constant. d/dx(ln(b))=0.
     
  6. Nov 6, 2008 #5
    Because when you use the product rule on
    d/dx (ln(b) * x) you get (1 + ln (b))
     
  7. Nov 6, 2008 #6

    Dick

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    d/dx(ln(x)*x)=(1+ln(x)). There's a b in there. Not two x's.
     
  8. Nov 6, 2008 #7
    so could you please tell me how to solve
    d/dx (ln(b) * x)
     
  9. Nov 6, 2008 #8

    Dick

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    d/dx(ln(b))=(d/dx(b))/b. If b is a constant then d/dx(b)=0, right? d/dx(ln(b)*x)=ln(b).
     
  10. Nov 6, 2008 #9
    Wow, I'm so stupid.

    Thanks a lot!!! You saved my day!!!
     
  11. Nov 6, 2008 #10

    Integral

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    ln(b) is a constant.

    [tex] \frac {d} {dx} ln(b) x = ln(b) \frac {dx} {dx} = ln(b) [/tex]
     
  12. Nov 6, 2008 #11
    look at the definition of derivatives and you will surely understand how does that fomula work
     
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