using diff func of a func..get from this to this???!!!!

Discussion in 'Undergraduate Math' started by Electromechy1, Sep 20, 2003.

  1. hi guys,
    came across this one in book i'm using, its in the diff func of a func section
    so no cheating using the standard derivative e^ax=ae^ax. how do i wtite this in
    standard form so i can use func of a func?? knowing my luck will come up on
    exam, so a solution would be appreciated.

    5e^2t+1 to 10e^2t+1

    best regards

    shaun
     
    Electromechy1, Sep 20, 2003
    #1
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  2. Electromechy1

    David Moran Guest

    I'll rewrite it as y=(5e^2t+1)^(10e^2t+1)

    First take the log of both sides and use the Power Rule.
    ln y=(10e^2t+1)ln(5e^2t+1)

    Now differentiate implicitly
    (1/y)(dy/dx)=(10e^2t+1)(10e^2t/(5e^2t+1))+(10e^2t)ln(5e^2t+1)
    dy/dx=(5e^2t+1)^(10e^2t+1)((10e^2t+1)(10e^2t/(5e^2t+1))+(10e^2t)ln(5e^2t+1)

    David Moran
     
    David Moran, Sep 20, 2003
    #2
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  3. Electromechy1

    Paul Sperry Guest

    ?

    What is "this"?
     
    Paul Sperry, Sep 21, 2003
    #3
  4. thanx david

    regards

    shaun
     
    Electromechy1, Sep 21, 2003
    #4
  5. Alternatively, you could just use the chain rule. You've got y = f(x)^g(x)
    =e^[ln(f(x))*g(x)], so
    dy/dx = e^[ln(f(x))*g(x)]*[1/f(x)*f'(x)*g(x)+ln(f(x))*g'(x)], and plug in
    what f(x) and g(x) are.

    Whatever you're more comfortable with. Objectively, it looks like six of
    one and half a dozen of the other to me.

    I was going to suggest that you need to do it both ways and compare the
    answers, but it's likely that it'll just be real hard to chase through all
    the algebra, and you'll be back for us to check your work. That's too ugly,
    so just take my word that the answers are really the same.

    Jon Miller
     
    Jonathan Miller, Sep 21, 2003
    #5
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