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Logarithmic differentiation

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  • 18-07-2012 12:17pm
    #1
    Inbox
    Closed Accounts Posts: 810 ✭✭✭


    y=xe^x^2/(3x+2)(x+4)

    I'm stuck on this bit.

    Ln(xe^x^2) does this become x^2ln(xe)?

    Thanks


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    #2
    TheBody
    Registered Users Posts: 5,619 ✭✭✭TheBody


    Inbox wrote: »
    y=xe^x^2/(3x+2)(x+4)

    I'm stuck on this bit.

    Ln(xe^x^2) does this become x^2ln(xe)?

    Thanks

    To begin you need to know the laws of logs.

    Anyway,

    [latex]ln(xe^{x^2})=ln(x)+ln(e^{x^2})[/latex]

    =[latex]ln(x)+x^2ln(e)[/latex]

    =[latex]ln(x)+x^2[/latex]

    Note: I haven't done ANY differentiating. Just borke down that statement that you wanted using the laws of logs.


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    #3
    TheBody
    Registered Users Posts: 5,619 ✭✭✭TheBody


    I just read your op again. Can I ask why you are using logarithmic differentiation? Just use the quotient rule!!


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    #4
    Inbox
    Closed Accounts Posts: 810 ✭✭✭Inbox


    TheBody wrote: »
    I just read your op again. Can I ask why you are using logarithmic differentiation? Just use the quotient rule!!

    Hi, that's what the question was and that was from a past exam paper i was doing.

    So the lecturer was wrong asking it that way. i will try using the quotient rule and see if i get the same answer.


  • Options
    #5
    Inbox
    Closed Accounts Posts: 810 ✭✭✭Inbox


    TheBody wrote: »
    To begin you need to know the laws of logs.

    Anyway,

    [latex]ln(xe^{x^2})=ln(x)+ln(e^{x^2})[/latex]

    =[latex]ln(x)+x^2ln(e)[/latex]

    =[latex]ln(x)+x^2[/latex]

    Note: I haven't done ANY differentiating. Just borke down that statement that you wanted using the laws of logs.

    Thanks , i only seem to have the basic rules. that's a great help.


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