implicit differentiation

**bigwave** · March 10th 2010, 05:11 PM

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:
$y' = \frac{4xy\sqrt{xy}-y}{x-2x^2\sqrt{xy}}$

**Archie Meade** · March 10th 2010, 05:32 PM

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.

**Black** · March 10th 2010, 05:32 PM

$\frac{1}{2}(xy)^{-1/2}(xy'+y)-(x^2y'+2xy)=0$

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

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

Multiply numerator and denominator by (xy)^1/2 to get the answer.

**bigwave** · March 10th 2010, 05:58 PM

thanks got it
just have to carefull

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