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Thread: [SOLVED] Partial derivation

  1. #1
    Newbie kirbyiwaki's Avatar
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    Unhappy [SOLVED] Partial derivation

    I'm new at this, so...
    $\displaystyle yln{x}\ln{y}dx+dy=0$

    then, making it a function:

    $\displaystyle f(x,y)=\frac{dy}{dx}= -yln{x}\ln{y} $

    $\displaystyle \frac{\partial{f}}{\partial{y}}= ?$

    any help (and further explanations) will be great...

    thanks in advance ^^
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  2. #2
    Moo
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    A Cute Angle Moo's Avatar
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    Hello,
    Quote Originally Posted by kirbyiwaki View Post
    I'm new at this, so...
    $\displaystyle yln{x}\ln{y}dx+dy=0$

    then, making it a function:

    $\displaystyle f(x,y)=\frac{dy}{dx}= -yln{x}\ln{y} $

    $\displaystyle \frac{\partial{f}}{\partial{y}}= ?$

    any help (and further explanations) will be great...

    thanks in advance ^^
    In other words, you're asked for the partial derivative of $\displaystyle -y \ln(x) \ln(y)$, with respect to y.

    This means that you have to consider x as a constant. Thus $\displaystyle \ln(x)$ will be treatead as a constant. If you can't see it, substitute it by k.

    $\displaystyle g(y)=-k \cdot y\ln(y)$

    Can you find $\displaystyle g'(y)$ ?
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  3. #3
    Newbie kirbyiwaki's Avatar
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    OMG, that was easy xD... looks like the $\displaystyle \ln$ scared me >_>... thanks n__n"
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