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Math Help - Implicit differentiation

  1. #1
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    Implicit differentiation

    Alright, so I am in calc 1 and I am learning implicit differentiation. I understand it more or less but I am still having difficulties. I have the problem √xy = x + y. The solution is y' = 2(√xy)-y / x-2(√xy). I don't get how they simplified to this point.
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  2. #2
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    Re: Implicit differentiation

    Hello, paperclip27!

    I must assume you know how to differentiate implicitly.


    Implicit differentiation: . \sqrt{xy} \:=\: x + y

    The solution is: . y' \:=\:\dfrac{2\sqrt{xy}-y}{x-2\sqrt{xy}}

    We have: . (xy)^{\frac{1}{2}} \:=\:x + y

    Then: . \tfrac{1}{2}(xy)^{-\frac{1}{2}}(y + xy') \:=\: 1 + y'

    . . . . . . . . . \frac{y + xy'}{2\sqrt{xy}} \:=\: 1 + y'

    . . . . . . . . . y + xy' \:=\:2\sqrt{xy}(1 + y')

    . . . . . . . . . y + xy' \:=\:2\sqrt{xy} + 2\sqrt{xy}\,\!y'

    . . . . . xy' - 2\sqrt{xy}\,\!y' \:=\:2\sqrt{xy} - y

    . . . . . (x - 2\sqrt{xy})y' \;=\;2\sqrt{xy} - y

    . . . . . . . . . . . . . y' \:=\:\frac{2\sqrt{xy} - y}{x - 2\sqrt{xy}}

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  3. #3
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    Re: Implicit differentiation

    That makes perfect sense! Thank you! I wasn't putting the y + xy' on top of the fraction and then not getting the rest of it.
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