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Math Help - evaluate y' at (2,1)

  1. #1
    Member mybrohshi5's Avatar
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    evaluate y' at (2,1)

    For the equation given below, evaluate at the point .



    y' at (2,1) = ?

    so i tried taking the natural log of everything then differentiating both sides and solving for y'.

    ln2 + lne^xy - ln2 + lnx = lny +ln9.78

    xy + lnx = lny + ln9.78

    (1/y)y' = y + xy' + 1/x + 0

    y' = y^2 + xy'y + y/x

    y' = y^2+(y/x) / 1-xy

    then if you plug in (2,1) i get -1.5 and thats not right =(

    any advice?
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  2. #2
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    Quote Originally Posted by mybrohshi5 View Post
    For the equation given below, evaluate at the point .



    y' at (2,1) = ?

    so i tried taking the natural log of everything then differentiating both sides and solving for y'.

    ln2 + lne^xy - ln2 + lnx = lny +ln9.78

    any advice? yes, the entire step above is invalid

    ln(a+b) does not equal ln(a) + ln(b)
    ...
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  3. #3
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    Also, you can't take the log of everything and then differentiate as you would be differentiating log(y) and not y as required..
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  4. #4
    Member mybrohshi5's Avatar
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    ok i guess i have not a clue what to do then.

    should i start off by taking the natural log of both sides?

    then differentiating with u'/u for both sides?

    then solving for y'?

    thank you
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  5. #5
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    why log? just take the derivative straight up.

    \frac{d}{dx}(2e^{xy} - 2x = y + k)

    2e^{xy}(xy' + y) - 2 = y'

    2xy'e^{xy} + 2ye^{xy} - 2 = y'

    2ye^{xy} - 2 = y' - 2xy'e^{xy}

    2ye^{xy} - 2 = y'(1 - 2xe^{xy})

    \frac{2(ye^{xy}-1)}{1 - 2xe^{xy}} = y'
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  6. #6
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    Quote Originally Posted by mybrohshi5 View Post
    ok i guess i have not a clue what to do then.

    should i start off by taking the natural log of both sides?

    then differentiating with u'/u for both sides?

    then solving for y'?

    thank you
    You shouldn't have to take logarithms. You can just differentiate implicity.

    2e^{xy}-2x=y+9.78

    The derivative of 2e^{xy} is kind of tricky. Set u=xy and differentiate as follows:

    2e^{u}\frac{du}{dx}

    \frac{du}{dx}=x\frac{dy}{dx}+y (by product rule)

    =2e^u(\frac{xdy}{dx}+y)

    =2e^{xy}(\frac{xdy}{dx}+y)

    So now that you know the derivative of this term, the rest should be obvious.
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