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Problem of Calculus: derivative, i guess logarithmic differentiation

  1. Jun 9, 2012 #1
    1. The problem statement, all variables and given/known data

    Image of the problem: http://prntscr.com/addkf

    2. Relevant equations

    My question is how I can solve the equation I gave above.

    Should I use logarithmic differentiation?

    Because I think that the logarithmic differentiation is used when y = (the equation) but my problem is f (x) = (the equation).

    3. The attempt at a solution

    Possible solutions: http://prntscr.com/addwu
     
  2. jcsd
  3. Jun 9, 2012 #2
    try using (f+g)'=f'+g', then use logarithmic differentiation for f and g separately.

    In your link to worked example, first option is wrong because you never took derivative in sense of a^(5x-1) for instance. Second option is wrong because Log does not distrubute across addition like that.
     
  4. Jun 10, 2012 #3
    What about this possible solution to the problem:([itex]x^{(5x-1)}[/itex][itex]\frac{5x-1}{x}[/itex]+5lnx) - 2x[itex]e^{(11-x^2)}[/itex]
     
  5. Jun 10, 2012 #4

    SammyS

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    Hello BlaZz. Welcome to PF !

    It really helps us, if you make your images visible in your post.

    attachment.php?attachmentid=48215&stc=1&d=1339372887.png

    attachment.php?attachmentid=48216&stc=1&d=1339372979.png

    Show us how you got that.

    The [itex]\displaystyle -2x\,e^{11-x^2}[/itex] is the derivative of [itex]\displaystyle e^{11-x^2}.[/itex]
     

    Attached Files:

    Last edited: Jun 10, 2012
  6. Jun 10, 2012 #5
    I attach the procedure for which i got the result, i need that someone say me if i am right or wrong.

    Thanks to all the members that comment in my thread and try to help me.
     

    Attached Files:

  7. Jun 10, 2012 #6

    SammyS

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    attachment.php?attachmentid=48226&d=1339384422.png
    Yes, that answer looks fine !
     
  8. Jun 12, 2012 #7

    SammyS

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    So, I gave the x5x-1 term to WolframAlpha to differentiate.

    WolframAlpha did not use logarithmic differentiation. Instead, it used the following scheme.
    Suppose we have a function that can be expressed in the following way,

    [itex]\displaystyle f(x)=g\left(u(x),\,v(x)\right)[/itex]

    then the derivative of f(x) can be obtained as follows.

    [itex]\displaystyle \frac{d}{dx}f(x)=\frac{d}{dx}g\left(u(x),\,v(x) \right)=\frac{\partial g(u,v)}{\partial u}\frac{d\,u(x)}{dx}+\frac{\partial g(u,v)}{\partial v}\frac{d\,v(x)}{dx}[/itex]
    For the case of [itex]\displaystyle f(x)=x^{5x-1}\,,[/itex] we have [itex]\displaystyle g(u,\,v)=u^v\,,[/itex] where [itex]\displaystyle u(x)=x\,,[/itex] and [itex]\displaystyle u(x)=5x-1\,.[/itex]

    ∂g/∂u treats the function g, as if the base of the expression, in this case x by itself, is the part with the variable, and treats the exponent as being constant, as is the case of a power function.

    ∂g/∂v treats the exponent of the expression, in this case x by itself, as being the part with the variable, and treats the base as being constant, as is the case of an exponential function.
     
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