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Thread: 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

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

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

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

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

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