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Thread: Hello, need help with finding equation for a tangent line with the given function

  1. #1
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    Hello, need help with finding equation for a tangent line with the given function

    Hello! Thank you for your time and I really need help with finding the equation of the tangent line for 1+ln2xy=e^(2x-y) at the point (1/2,1)

    If someone would please help me and if they could add steps to it as well so that I can have a better understanding. Its probably simple and I'm just overthinking but I could really use some help
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    Re: Hello, need help with finding equation for a tangent line with the given function

    Quote Originally Posted by Sonofrhot6 View Post
    Hello! Thank you for your time and I really need help with finding the equation of the tangent line for 1+ln2xy=e^(2x-y) at the point (1/2,1)

    If someone would please help me and if they could add steps to it as well so that I can have a better understanding. Its probably simple and I'm just overthinking but I could really use some help
    In a few words: take the implicit derivative of both sides:
    $\displaystyle \frac{1}{2xy} \cdot \left ( 2y + 2x \cdot \frac{dy}{dx} \right ) = \left ( 2 - \frac{dy}{dx} \right ) \cdot e^{2x - y}$

    Usually I would recommend solving for dy/dx first, but in this case it may be simpler for you to put in the coordinates of the point, then solve for dy/dx.

    -Dan
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    Re: Hello, need help with finding equation for a tangent line with the given function

    $1+\ln(2xy) = e^{2x-y}$

    differentiate by $x$ on each side, $\dfrac{dy}{dx} = y^\prime$

    $\dfrac{2y+2xy^\prime}{2xy} = e^{2x-y}(2-y^\prime)$

    Solve for $y^\prime$

    $y^\prime = -\dfrac{y \left(e^y-2 e^{2 x} x\right)}{x \left(e^{2 x} y+e^y\right)}$

    now just plug your point $\left(\dfrac 1 2,~1\right)$ and evaluate

    I imagine your prof will want more details. I leave it to you to provide them.
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