# (xy^2-lny-(x+y)/x=0)'

• Aug 5th 2013, 08:00 AM
Boo
(xy^2-lny-(x+y)/x=0)'
Please, what is a derivative of:
$\displaystyle (xy^2-lny-\frac{x+y}{x}=0)'$ ?
Many thanks!
• Aug 5th 2013, 08:19 AM
Shakarri
Re: (xy^2-lny-(x+y)/x=0)'
What are you differentiating with respect to? Do you know what a partial derivative is?
• Aug 5th 2013, 11:20 AM
Soroban
Re: (xy^2-lny-(x+y)/x=0)'
Hello, Boo!

Quote:

$\displaystyle xy^2- \ln y-\frac{x+y}{x}\:=\:0$
$\displaystyle \text{Find }y'$

Multiply by $\displaystyle x\!:\;x^2y^2 - x\ln y - x - y \:=\:0$

Differentiate implicitly:

. . $\displaystyle x^2\!\cdot\!2yy' + 2x\!\cdot\!y^2 - x\!\cdot\!\frac{y'}{y} + 1\!\cdot\!\ln y - 1 - y' \:=\:0$

. . $\displaystyle 2x^2yy' + 2xy^2 - \frac{x}{y}y' - \ln y - 1 - y' \:=\:0$

. . $\displaystyle 2x^2yy' - \frac{x}{y}y' - y' \:=\:1 - 2xy^2 + \ln y$

. . $\displaystyle \left(2x^2y - \frac{x}{y} - 1\right)y' \:=\:1-2xy^2+ \ln y$

Multiply by $\displaystyle y\!:\;\left(2x^2y^2 - x - y\right)y' \:=\:y(1-2xy^2+ \ln y)$

Therefore: .$\displaystyle y' \;=\;\frac{y(1-2xy^2+ \ln y)}{2x^2y^2 - x - y}$
• Aug 6th 2013, 06:14 AM
Boo
Re: (xy^2-lny-(x+y)/x=0)'
Please, can u explain me first line, differentiate implicitly?
Many thanks!
• Aug 6th 2013, 08:05 AM
Soroban
Re: (xy^2-lny-(x+y)/x=0)'
Hello, Boo!

Quote:

Please, can u explain me first line, "differentiate implicitly"? . No!

If you do not understand Implicit Differentiation,
. . you have no business attempting this problem.