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Math Help - Implicit Differentiation question

  1. #1
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    Implicit Differentiation question

    Hi
    Need help to use the Implicit Differentiation method on  \frac{xy}{lnx}.
    I have tried many ways to get the right answer. Below is what i have done:

    =\frac{\frac{dy}{dx}lnx-\frac{xy}{x}}{(lnx)^2}

    \frac{\frac{dy}{dx}lnx}{(lnx)^2}-\frac{\frac{xy}{x}}{(lnx)^2}


    \frac{dy}{dx}\frac{1}{lnx}=\frac{\frac{xy}{x}}{(ln  x)^2}

    \frac{dy}{dx}=\frac{y}{(ln)^2}-\frac{1}{lnx}

    \frac{dy}{dx}=\frac{y-lnx}{(lnx)^2}

    Book's answer is \frac{y-ylnx}{xlnx}
    P.S
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  2. #2
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    Quote Originally Posted by Paymemoney View Post
    Hi
    Need help to use the Implicit Differentiation method on  \frac{xy}{lnx}.
    I have tried many ways to get the right answer. Below is what i have done:

    =\frac{\frac{dy}{dx}lnx-\frac{xy}{x}}{(lnx)^2}

    \frac{\frac{dy}{dx}lnx}{(lnx)^2}-\frac{\frac{xy}{x}}{(lnx)^2}


    \frac{dy}{dx}\frac{1}{lnx}=\frac{\frac{xy}{x}}{(ln  x)^2}

    \frac{dy}{dx}=\frac{y}{(ln)^2}-\frac{1}{lnx}

    \frac{dy}{dx}=\frac{y-lnx}{(lnx)^2}

    Book's answer is \frac{y-ylnx}{xlnx}
    P.S
    You should apply the product rule when you differentiate xy.
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  3. #3
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    ok i'll try that
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  4. #4
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    ok after doing it again this is what i got:
    \frac{xy}{lnx}=\frac{y+x}{lnx}

    \frac{dy}{dx}lnx-\frac{y+x}{x}

    \frac{\frac{dy}{dx}xlnx}{x}-\frac{y+x}{x}

    \frac{\frac{dy}{dx}xlnx-y+x}{x(lnx)^2}

    \frac{\frac{dy}{dx}lnx}{x(lnx)^2}=\frac{y+x}{x(lnx  )^2}

    \frac{y+x-x{lnx}}{x(lnx)^2}
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  5. #5
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    Quote Originally Posted by Paymemoney View Post
    ok after doing it again this is what i got:
    \frac{xy}{lnx}=\frac{y+x}{lnx}
    What is this ?
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  6. #6
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    Quote Originally Posted by Paymemoney View Post
    ok after doing it again this is what i got:
    \frac{xy}{lnx}=\frac{y+x}{lnx}

    \frac{dy}{dx}lnx-\frac{y+x}{x}

    \frac{\frac{dy}{dx}xlnx}{x}-\frac{y+x}{x}

    \frac{\frac{dy}{dx}xlnx-y+x}{x(lnx)^2}

    \frac{\frac{dy}{dx}lnx}{x(lnx)^2}=\frac{y+x}{x(lnx  )^2}

    \frac{y+x-x{lnx}}{x(lnx)^2}
    \left(\frac{xy}{\ln x}\right)' =\frac{(y + xy')\ln x - xy(\frac{1}{x})} {\ln^2 x}= 0

    (y + xy')\ln x - y= 0

    (y + xy') = \frac{y}{\ln x}

    xy' = \frac{y}{\ln x} - y

    y' = \frac{y}{x\ln x} - \frac{y}{x}

    y' = \frac{y - y\ln x}{x\ln x}
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  7. #7
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    Quote Originally Posted by Paymemoney View Post
    Hi
    Need help to use the Implicit Differentiation method on  \frac{xy}{lnx}.
    I have tried many ways to get the right answer. Below is what i have done:

    =\frac{\frac{dy}{dx}lnx-\frac{xy}{x}}{(lnx)^2}

    \frac{\frac{dy}{dx}lnx}{(lnx)^2}-\frac{\frac{xy}{x}}{(lnx)^2}


    \frac{dy}{dx}\frac{1}{lnx}=\frac{\frac{xy}{x}}{(ln  x)^2}

    \frac{dy}{dx}=\frac{y}{(ln)^2}-\frac{1}{lnx}

    \frac{dy}{dx}=\frac{y-lnx}{(lnx)^2}

    Book's answer is \frac{y-ylnx}{xlnx}
    P.S
    What is  \frac{xy}{lnx} equal to???
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  8. #8
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    Quote Originally Posted by General View Post
    What is this ?
    i used the product rule on xy to get y+x
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  9. #9
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    omg i forgot to add that \frac{xy}{lnx}=2
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  10. #10
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    Quote Originally Posted by Paymemoney View Post
    omg i forgot to add that \frac{xy}{lnx}=2
    1- You did not apply the product rule on xy correctly.
    2- Its wrong to apply the product rule on the numerator first, then differentiate the whole ratio.
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  11. #11
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    In order to use implicit differentiation to find dy/dx, you have to have an equation to start with.

    So I ask what xalk asked before: "What is <br />
\frac{xy}{lnx}<br />
equal to?"
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  12. #12
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    Quote Originally Posted by Paymemoney View Post
    Hi
    Need help to use the Implicit Differentiation method on  \frac{xy}{lnx}.
    I have tried many ways to get the right answer. Below is what i have done:

    =\frac{\frac{dy}{dx}lnx-\frac{xy}{x}}{(lnx)^2}

    \frac{\frac{dy}{dx}lnx}{(lnx)^2}-\frac{\frac{xy}{x}}{(lnx)^2}


    \frac{dy}{dx}\frac{1}{lnx}=\frac{\frac{xy}{x}}{(ln  x)^2}

    \frac{dy}{dx}=\frac{y}{(ln)^2}-\frac{1}{lnx}

    \frac{dy}{dx}=\frac{y-lnx}{(lnx)^2}

    Book's answer is \frac{y-ylnx}{xlnx}
    P.S
    Another way of doing it is the following:

    \frac{xy}{lnx}=2\Longrightarrow xy = 2lnx \Longrightarrow \frac{d(xy)}{dx} =2\frac{d(lnx)}{dx} \Longrightarrow y+xy' = \frac{2}{x} \Longrightarrow y' = \frac{2-xy}{x^2}.

    And if we substitute 2 with xy/lnx we have:

     y' = \frac{y-ylnx}{lnx}
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