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Thread: more implict diff.

  1. #1
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    more implict diff.

    Ok, I don't know how to go about this one at all.

    Find dy/dx by implicit differentiation.
    e^(x/y) = 8x - y
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  2. #2
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    Quote Originally Posted by kwivo View Post
    Find dy/dx by implicit differentiation.
    e^(x/y) = 8x - y
    $\displaystyle \left(\frac xy\right)'\cdot e^{x/y}=8-y'$

    Find $\displaystyle y'$
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  3. #3
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    I got dy/dx= ye^(x/y) + 8y^2/ xe^(x/y) + y^2

    Is that right because the hw prog is telling me it's wrong. Maybe I didn't simplify it all the way yet?
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  4. #4
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    Quote Originally Posted by kwivo View Post
    Ok, I don't know how to go about this one at all.

    Find dy/dx by implicit differentiation.
    e^(x/y) = 8x - y
    Quote Originally Posted by Krizalid View Post
    $\displaystyle \left(\frac xy\right)'\cdot e^{x/y}=8-y'$

    Find $\displaystyle y'$
    Quote Originally Posted by kwivo View Post
    I got dy/dx= ye^(x/y) + 8y^2/ xe^(x/y) + y^2

    Is that right because the hw prog is telling me it's wrong. Maybe I didn't simplify it all the way yet?
    $\displaystyle \frac{d}{dx} \frac{x}{y} = \frac{1 \cdot y - xy^{\prime}}{y^2}$

    So from Krizalid's post:
    $\displaystyle \left ( \frac{y - xy^{\prime}}{y^2} \right ) \cdot e^{x/y}=8-y^{\prime}$

    $\displaystyle y e^{x/y} - xy^{\prime}e^{x/y} = 8y^2 - y^2y^{\prime}$

    $\displaystyle y^2y^{\prime} - xy^{\prime}e^{x/y} = 8y^2 - y e^{x/y}$

    $\displaystyle (y^2 - xe^{x/y})y^{\prime} = 8y^2 - y e^{x/y}$

    $\displaystyle y^{\prime} = \frac{8y^2 - y e^{x/y}}{y^2 - xe^{x/y}}$

    (And you really need to use parenthesis!)

    -Dan
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