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Math Help - Calculus Implicit Differentiation

  1. #1
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    Question Calculus Implicit Differentiation

    Use implicit differentiation to find dy/dx if cos xy = 2x^2-3y


    Thanks!
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  2. #2
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    Quote Originally Posted by SoYah View Post
    Use implicit differentiation to find dy/dx if cos xy = 2x^2-3y


    Thanks!
     \frac{d}{dx} xy = \frac{d}{dx} (2x^2-3y)

     y + x\frac{dy}{dx} =  4x-3\frac{dy}{dx}

      x\frac{dy}{dx}+3\frac{dy}{dx} =  4x-y

      \frac{dy}{dx}(x+3) =  4x-y

      \frac{dy}{dx} =  \frac{4x-y}{x+3}
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  3. #3
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    Hello, SoYah!

    Use implicit differentiation to find \tfrac{dy}{dx} if \cos(xy) \:= \:2x^2-3y

    We have: . -\sin(xy)\left(x\frac{dy}{dx} + y\right) \:=\:4x - 3\frac{dy}{dx}

    . . . . . . -x\sin(xy)\frac{dy}{dx} - y\sin(xy) \:=\:4x - 3\frac{dy}{dx}

    . . . . . . . . . 3\frac{dy}{dx} - x\sin(xy)\frac{dy}{dx} \:=\:4x + y\sin(xy)


    Factor: . \bigg[3-x\sin(xy)\bigg]\frac{dy}{dx} \:=\:4x + y\sin(xy)


    Therefore: . \frac{dy}{dx} \:=\:\frac{4x+y\sin(xy)}{3-x\sin(xy)}

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  4. #4
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    Quote Originally Posted by Mush View Post
     \frac{d}{dx} xy = \frac{d}{dx} (2x^2-3y)

     y + x\frac{dy}{dx} = 4x-3\frac{dy}{dx}

     x\frac{dy}{dx}+3\frac{dy}{dx} = 4x-y

     \frac{dy}{dx}(x+3) = 4x-y

     \frac{dy}{dx} = \frac{4x-y}{x+3}
    You missed the cos on the left hand side
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  5. #5
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    Quote Originally Posted by mr fantastic View Post
    You missed the cos on the left hand side
    Ahhh, so I did.
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