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Thread: Implicit differentiation

  1. #1
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    Implicit differentiation

    x^3y^4-9xsiny=14
    I took the derivitives as
    (3x^2y^4)+((x^3)4y^3(y`))-(9siny+9xcosy(y`))
    I solved for y` and got
    y` =(3x^2y^4+9siny)/x^3(4y^3)-9cosy)
    But this doesn't come out as correct.

    Any suggestions?
    Thanks
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  2. #2
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    Hello, juventinoalex!

    $\displaystyle x^3y^4 - 9x\sin y \:=\:14$

    $\displaystyle \text{I differentiated: }\;3x^2y^4 + x^34y^3(y')-\bigg(9\sin y+9x\cos y(y')\bigg) \:=\:0$ . Yes

    $\displaystyle \text{I solved for }y'\text{ and got: }\;y' \:=\:\dfrac{3x^2y^4 + 9\sin y}{4x^3y^3-9\cos y}$ . No

    $\displaystyle \text{But this doesn't come out as correct.}$

    We have: .$\displaystyle 3x^2y^4 - 4x^3y^3(y') - 9\sin y - 9x\cos y (y') \;=\;0$

    . . .Then: .$\displaystyle 4x^3y^3(y') + 9x\cos y(y') \;=\;3x^2y^4 - 9\sin y $

    . . Factor: .$\displaystyle (4x^3y^3 + 9x\cos y)(y') \;=\;3x^2y^4 - 9\sin y$


    Therefore: .$\displaystyle y' \;=\;\dfrac{3x^2y^4 - 9\sin y}{4x^3y^3 + 9x\cos y} $
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