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Math Help - Implicit natural log differentiation

  1. #1
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    Implicit natural log differentiation

    How would i go about the steps of solving for dy/dx implicitly

    \ln x^4y^4 = x^3 + y^3
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  2. #2
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    Quote Originally Posted by x5pyd3rx View Post
    How would i go about the steps of solving for dy/dx implicitly

    \ln x^4y^4 = x^3 + y^3

    \frac{4x^3}{x^4}y^4 + \ln(x^4)(4y^3)\frac{dy}{dx}=3x^2+3y^2\frac{dy}{dx}
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  3. #3
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    Quote Originally Posted by vince View Post
    \frac{4x^3}{x^4}y^4 + \ln(x^4)(4y^3)\frac{dy}{dx}=3x^2+3y^2\frac{dy}{dx}
    He has given you the solution, but not much of an explanation. Let's look just at \ln(x^4)(y^4)

    This is a product rule so we need to use f'g+fg'
    f=\ln(x^4)
    f'=\frac{1}{x^4}*{4x^3} This is a chain rule using the fact that the derivative of \ln(x) = \frac{1}{x}

    g=y^4
    For g' think about y as a function of x. In other words, if we take a hypothetical y=(x^2+1)^4, the derivative would be 4(x^2+1)^3*2x. This is precisely g'=4y^3*\frac{dy}{dx}
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