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Math Help - Derivatives of equation using ln and e

  1. #1
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    Derivatives of equation using ln and e

    Here is the equation:

    Find \frac{dy}{dx} for e^ylnx+xy^2=y

    I know that

    f' of e^y=e^y

    f' of lnx=\frac{1}{x}

    and

    f' of xy^2=2xy

    I suppose I use the chain rule to work out the answer?

    This is part of an assignment that is reviewing a few things we've covered and earlier today I listed some other problems, I have a multiple choice list of solutions that I'm posting for reference

    a.) \frac{-(e^y+xy^2)}{x(e^ylnx+2y-1)}

    b.) \frac{-(e^y+xy)}{x(e^ylnx+2xy-1)}

    c.) \frac{e^y+xy}{e^ylnx+2xy}

    d.) None of these

    Thanks in advance for your help. I'm just a bit intimidated by the possibility of making a mistake in the steps
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Here you go

    Quote Originally Posted by XIII13Thirteen View Post
    Here is the equation:

    Find \frac{dy}{dx} for e^ylnx+xy^2=y

    I know that

    f' of e^y=e^y

    f' of lnx=\frac{1}{x}

    and

    f' of xy^2=2xy

    I suppose I use the chain rule to work out the answer?

    This is part of an assignment that is reviewing a few things we've covered and earlier today I listed some other problems, I have a multiple choice list of solutions that I'm posting for reference

    a.) \frac{-(e^y+xy^2)}{x(e^ylnx+2y-1)}

    b.) \frac{-(e^y+xy)}{x(e^ylnx+2xy-1)}

    c.) \frac{e^y+xy}{e^ylnx+2xy}

    d.) None of these

    Thanks in advance for your help. I'm just a bit intimidated by the possibility of making a mistake in the steps
    e^{y}\cdot\ln(x)+xy^2=y


    then \frac{e^{y}}{x}+\ln(x)e^{y}\cdot{y'}+2xyy'+y^2=y'...now solve for y'
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  3. #3
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    Quote Originally Posted by Mathstud28 View Post
    e^{y}\cdot\ln(x)+xy^2=y


    then \frac{e^{y}}{x}+\ln(x)e^{y}\cdot{y'}+2xyy'+y^2=y'...now solve for y'
    I'm a bit lost. Inserting the y' into the equation looks familiar, but I think I've forgotten how this principle is applied
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  4. #4
    MHF Contributor kalagota's Avatar
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    Taguig City, Philippines
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    just note that y' = \frac{dy}{dx}

    it came out because of your "chain rule"..

    if your y=e^y \ln{x} + xy^2 and you are looking for y' = \frac{dy}{dx}.. here it is..

    taking the derivative of both sides yields to
    \frac{dy}{dx} = \underbrace{\frac{e^y}{x} + e^y\, \ln{x} \, \frac{dy}{dx}}_{product \, rule \, and \, chain \, rule \, on \, e^y} + \underbrace{y^2 + 2xy \, \frac{dy}{dx}}_{product \, rule \, and \, chain \, rule \, on \, y^2}
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  5. #5
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    Quote Originally Posted by kalagota View Post
    just note that y' = \frac{dy}{dx}

    it came out because of your "chain rule"..

    if your y=e^y \ln{x} + xy^2 and you are looking for y' = \frac{dy}{dx}.. here it is..

    taking the derivative of both sides yields to
    \frac{dy}{dx} = \underbrace{\frac{e^y}{x} + e^y\, \ln{x} \, \frac{dy}{dx}}_{product \, rule \, and \, chain \, rule \, on \, e^y} + \underbrace{y^2 + 2xy \, \frac{dy}{dx}}_{product \, rule \, and \, chain \, rule \, on \, y^2}
    Ahh, thank you for clarifying. It seems that the answer is "none of these" in this case. I suppose I assumed my answer would be very close to the answers provided, so I was wondering about the denominator in particular and the negative symbols. I guess the red herring was that there was no red herring
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