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Logarithmic differentiation

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  • 20-09-2015 8:38pm
    #1
    Hedgecutter
    Registered Users Posts: 1,030 ✭✭✭


    Here's a question I'm trying at the moment. The blue writhing is my work and red is the answer from the book.

    I don't understand where they get 6/(2x+1), I got 3/(2x+1).


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    #2
    Pherekydes
    Registered Users Posts: 16,120 ✭✭✭✭Pherekydes


    Don't forget the derivative of the bracket.


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    #3
    Hedgecutter
    Registered Users Posts: 1,030 ✭✭✭Hedgecutter


    Pherekydes wrote: »
    Don't forget the derivative of the bracket.

    Derivative of 2x+1 is 2, do I multiply that by 3 = 6 ?


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    #4
    Pherekydes
    Registered Users Posts: 16,120 ✭✭✭✭Pherekydes


    Yes.


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    #5
    Hedgecutter
    Registered Users Posts: 1,030 ✭✭✭Hedgecutter


    Pherekydes wrote: »
    Yes.



    So do you always work out the derivative of what's in the brackets (under the line) and multiply by what's above the line?


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    #6
    Moody_mona
    Registered Users Posts: 1,107 ✭✭✭Moody_mona


    The chain rule is being used here. The differentiation of log(f(x)) = (1/f(x)) times f'(x)


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    #7
    Yakuza
    Registered Users Posts: 5,141 ✭✭✭Yakuza


    As Mona above says, it's the Chain rule : dy/dx = dy/du * du/dx.
    y = ln(2x+1), set u = 2x+1, du/dx = 2
    y = ln(u) so dy/du = 1/u
    dy/dx = 1/u * 2 or 2/(2x+1)

    In the original question the function is 3 * ln(2x+1), so the answer is 3 * (2 / (2x +1) (by the product rule - d (uv)/dx = v du/dx + u dv/dx but one of the terms is a constant so its derivative is zero)


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