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Math Help - Derivative

  1. #1
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    Derivative

    Hello.

    Could someone explain me the following in the subject of derivative.

    y = ln x ; y'= 1/x

    Following that rule why doesn't y =ln x^5 equal y' = 1/x^5 but instead y' = 5/x
    If someone could explain me this I would be very greatful.

    PS! I am not an english speaking person so my english is limited.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by Kurumax View Post
    Following that rule why doesn't y =ln x^5 equal y' = 1/x^5 but instead y' = 5/x
    As a consequence of the Chainīs Rule:

    f(x)=\ln u(x)\Rightarrow f'(x)=\dfrac{1}{u(x)}\cdot u'(x)


    Fernando Revilla
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  3. #3
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    Thank you!
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  4. #4
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    I have a new question. How can i solve this derivative?

    y = ln 1/x^2

    I've been messing with this for some time now and the solution i got was y' = -1/x^2 but i do not believe it's right.
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  5. #5
    MHF Contributor harish21's Avatar
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    The question you are asking here is quite similar to your first question..

    Look at FernandoRevilla's suggestion in post #2

    f(x)=\ln (u(x)) where u(x)=\dfrac{1}{x^2} and then find the derivative... show your work f you are stuck...
    Last edited by mr fantastic; January 24th 2011 at 01:18 AM.
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  6. #6
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     y=ln \dfrac{1}{x^2} = -ln x^2

     y' = -\dfrac{1}{x^2} \cdot \dfrac{2x}{1} = -\dfrac{2}{x}

    Is this correct?
    Last edited by mr fantastic; January 24th 2011 at 01:19 AM.
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  7. #7
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by Kurumax View Post
    Sorry about this thread being in the wrong section. My english math knowladge is not so good.

     y=ln \dfrac{1}{x^2} = -ln x^2

     y' = -\dfrac{1}{x^2} \cdot \dfrac{2x}{1} = -\dfrac{2}{x}

    Is this correct?


    EDIT: Yes .. it is correct..
    Last edited by harish21; January 23rd 2011 at 10:26 AM. Reason: didnt read the answer carefully
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  8. #8
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    Thank you for confirming the answer.
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  9. #9
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    Quote Originally Posted by Kurumax View Post
    Sorry about this thread being in the wrong section. My english math knowladge is not so good.

     y=ln \dfrac{1}{x^2} = -ln x^2

     y' = -\dfrac{1}{x^2} \cdot \dfrac{2x}{1} = -\dfrac{2}{x}

    Is this correct?
    Correct.

    An even easier alternative: \displaystyle y = \ln{\left(\frac{1}{x^2}\right)} = \ln{(x^{-2})} = -2\ln{x}.

    So \displaystyle \frac{dy}{dx} = -\frac{2}{x}
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  10. #10
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    Thanks again everyone. Passed my test thanks to this help.
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