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Math Help - Logarithmic Differentiation Question

  1. #1
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    Logarithmic Differentiation Question

    Hi,

    I have been given the question differentiate the following equation logarithmically:

    Logarithmic Differentiation Question-mathq.jpg

    However, I don't yet understand factorising very well, and the answer given for this question has been factorised so I can't compare it with my own answer.

    I took the following steps to work this out:

    Logarithmic Differentiation Question-mathq2.jpg


    If someone can verify wherether this is correct or not, and tell me how if it's not, that would really useful.

    Thanks.
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  2. #2
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    Where did the 2 come from in your last line of working?

    \displaystyle \frac{e^x}{x}+e^x\ln x \neq \frac{2e^x}{x}+\ln x
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  3. #3
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    It's all correct except for line 7, where e^x is a factor.
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  4. #4
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    \displaystyle y=\exp{(e^x\ln{x})}\Rightarrow \frac{dy}{dx}=\exp{(e^x\ln{x})}\left[e^x\ln{x}+e^x*\frac{1}{x}\right]

    \displaystyle\Rightarrow x^{e^x}e^x\left[\ln{x}+\frac{1}{x}\right]
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  5. #5
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    Ah I see, I tried factorising a little so the \displaystyle \frac{e^x}{x}+e^x became \displaystyle \frac{2e^x}{x}

    I realise this should be \displaystyle \frac{2e^x * ln(x)}{x}

    and not

    \displaystyle \frac{2e^x}{x} + ln(x)

    When you say e^x is a factor on line 7, which e^x is this?
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  6. #6
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    You see why that isn't true though right? \displaystyle \frac{e^x}{x}+e^x \ne \frac{2e^x}{x} You can factor it however to \displaystyle e^x \left( \frac{1}{x}+\ln(x) \right)
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  7. #7
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    Ok I can see how that part is wrong now.

    If I replaced step 7 with this:


    Would it then be correct? I realise it can be factorised but, for now I'd rather work longhand and tackle factorising another day.
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  8. #8
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    Yes
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  9. #9
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    Brilliant thanks.
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