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

Finddy/dxby implicit differentiation.

e^(x/y)= 8x-y

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- Oct 9th 2007, 06:35 PMkwivomore implict diff.
Ok, I don't know how to go about this one at all.

Find*dy*/*dx*by implicit differentiation.

*e^(x/y)*= 8*x*-*y* - Oct 9th 2007, 06:39 PMKrizalid
- Oct 9th 2007, 07:12 PMkwivo
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? - Oct 10th 2007, 04:10 AMtopsquark
$\displaystyle \frac{d}{dx} \frac{x}{y} = \frac{1 \cdot y - xy^{\prime}}{y^2}$

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