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Math Help - Are these derivatives correct?

  1. #1
    Junior Member
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    Are these derivatives correct?

    Find the derivative of (xy+1)^3 = x - y^2 + 8 using implicit differentiation. Do not simplify.
    (xy+1)^3 = x - y^2 + 8

    3(xy+1)^2 \cdot (y + x \cdot \frac{dy}{dx}) = 1 - 2y \cdot \frac{dy}{dx}

    y + x \cdot \frac{dy}{dx} + \frac{2y \cdot \frac{dy}{dx}}{3(xy+1)^2} = \frac {1}{3(xy+1)^2}

    x \cdot \frac{dy}{dx} + \frac{2y \cdot \frac{dy}{dx}}{3(xy+1)^2} = \frac {1}{3(xy+1)^2} -y

    \frac{dy}{dx}[x + \frac{2y}{3(xy+1)^2}] = \frac {1}{3(xy+1)^2} -y

    \frac{dy}{dx} = \frac {\frac {1}{3(xy+1)^2} -y}{x + \frac{2y}{3(xy+1)^2}}

    The question stated not to simplify, so this would be my final answer.


    Find the derivative of \sin(xy) + x = e^y using implicit differentiation. Do not simplify.
    \sin(xy) + x = e^y

    \cos(xy) \cdot (y + x \cdot \frac{dy}{dx}) + 1 = e^y \cdot \frac{dy}{dx}

    \cos(xy) \cdot (y + x \cdot \frac{dy}{dx}) - e^y \cdot \frac{dy}{dx} = -1

     x \cdot \frac{dy}{dx} - \frac{e^y \cdot \frac{dy}{dx}}{\cos(xy)} = \frac{-1}{\cos(xy)} - y

     \frac{dy}{dx}[x - \frac{e^y}{\cos(xy)}] = \frac{-1}{\cos(xy)} - y

     \frac{dy}{dx} = \frac{\frac{-1}{\cos(xy)} - y}{[x - \frac{e^y}{\cos(xy)}]}

    I opted to not use logarithmic differentiation for some reason. Is this solution still valid?


    Thanks a lot!
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  2. #2
    Senior Member
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    Clarksville, ARk
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    Re: Are these derivatives correct?

    They look fine to me.

    I don't think that logarithmic differentiation would have helped .
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  3. #3
    Junior Member
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    Re: Are these derivatives correct?

    Thanks for the response!

    The other people in my class that I talked to said that they used logarithmic differention; I haven't tried it yet with this one yet.

    Though, as long as they're correct, that's all that matters at the moment.
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