# more implict diff.

• Oct 9th 2007, 06:35 PM
kwivo
more implict diff.

Find dy/dx by implicit differentiation.
e^(x/y) = 8x - y
• Oct 9th 2007, 06:39 PM
Krizalid
Quote:

Originally Posted by kwivo
Find dy/dx by implicit differentiation.
e^(x/y) = 8x - y

$\displaystyle \left(\frac xy\right)'\cdot e^{x/y}=8-y'$

Find $\displaystyle y'$
• Oct 9th 2007, 07:12 PM
kwivo
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 AM
topsquark
Quote:

Originally Posted by kwivo

Find dy/dx by implicit differentiation.
e^(x/y) = 8x - y

Quote:

Originally Posted by Krizalid
$\displaystyle \left(\frac xy\right)'\cdot e^{x/y}=8-y'$

Find $\displaystyle y'$

Quote:

Originally Posted by kwivo
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