# implicit differentiation

• Mar 10th 2010, 05:11 PM
bigwave
implicit differentiation
find dy/dx of $\sqrt{xy} = 1 + x^2y$

rewrite
$(xy)^{1/2}-x^2y = 1$
derivative of terms
$\frac{1}{2}(xy)^{-1/2}(xy' + y) - (x^2y' + y2x) = 0$

$\frac{xy'}{2\sqrt{xy}}+\frac{y}{2\sqrt{xy}}-x^2y'-2xy=0$

took more steps but did not get to the answer which is:
$y' = \frac{4xy\sqrt{xy}-y}{x-2x^2\sqrt{xy}}$
• Mar 10th 2010, 05:32 PM
Quote:

Originally Posted by bigwave
find dy/dx of $\sqrt{xy} = 1 + x^2y$

rewrite
$(xy)^{1/2}-x^2y = 1$
derivative of terms
$\frac{1}{2}(xy)^{-1/2}(xy' + y) - (\color{red}x^2\color{black}y' + y2x) = 0$

$\frac{xy'}{2\sqrt{xy}}+\frac{y}{2\sqrt{xy}}-\color{red}x^2\color{black}y'-2xy=0$

took more steps but did not get to the answer which is:
$y' = \frac{4xy\sqrt{xy}-y}{x-2x^2\sqrt{xy}}$

hi Bigwave,

you had an error in an early step.
• Mar 10th 2010, 05:32 PM
Black
$\frac{1}{2}(xy)^{-1/2}(xy'+y)-(x^2y'+2xy)=0$

$(x(xy)^{-1/2}-2x^2)y'=4xy-y(xy)^{-1/2}$

$y'=\frac{4xy-y(xy)^{-1/2}}{x(xy)^{-1/2}-2x^2}$.

Multiply numerator and denominator by (xy)^1/2 to get the answer.
• Mar 10th 2010, 05:58 PM
bigwave
thanks got it
thanks got it
just have to carefull(Cool)