1. ## Implicit differentiation

Alright, so I am in calc 1 and I am learning implicit differentiation. I understand it more or less but I am still having difficulties. I have the problem √xy = x + y. The solution is y' = 2(√xy)-y / x-2(√xy). I don't get how they simplified to this point.

2. ## Re: Implicit differentiation

Hello, paperclip27!

I must assume you know how to differentiate implicitly.

Implicit differentiation: . $\sqrt{xy} \:=\: x + y$

The solution is: . $y' \:=\:\dfrac{2\sqrt{xy}-y}{x-2\sqrt{xy}}$

We have: . $(xy)^{\frac{1}{2}} \:=\:x + y$

Then: . $\tfrac{1}{2}(xy)^{-\frac{1}{2}}(y + xy') \:=\: 1 + y'$

. . . . . . . . . $\frac{y + xy'}{2\sqrt{xy}} \:=\: 1 + y'$

. . . . . . . . . $y + xy' \:=\:2\sqrt{xy}(1 + y')$

. . . . . . . . . $y + xy' \:=\:2\sqrt{xy} + 2\sqrt{xy}\,\!y'$

. . . . . $xy' - 2\sqrt{xy}\,\!y' \:=\:2\sqrt{xy} - y$

. . . . . $(x - 2\sqrt{xy})y' \;=\;2\sqrt{xy} - y$

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

3. ## Re: Implicit differentiation

That makes perfect sense! Thank you! I wasn't putting the y + xy' on top of the fraction and then not getting the rest of it.