Im having trouble with an implicit differentiation problem.
Okay, x^3y^3-y=x
I started off with: x^3(3y^2y')+y^3(3x^2)-y'=1..
That would be correct, right? Also, with the x on the other side of the equal sign would be one throughout the whole thing?

Question

Im having trouble with an implicit differentiation problem. 
Okay, x^3y^3-y=x 
I started off with: x^3(3y^2y')+y^3(3x^2)-y'=1.. 
That would be correct, right? Also, with the x on the other side of the equal sign would  be one throughout the whole thing?

anonymous · Answer

That seems ok so far.

anonymous · Answer

I hope you don't mean to think that x=1.

anonymous · Answer

I thought we take the derivative of both sides? Like d|dx(x)=1?

anonymous · Answer

Or do I move that x to the lefthand side of the equation?

anonymous · Answer

No, that's fine.
What you have is correct so far, but you still need to solve that equation for y'

anonymous · Answer

The answer I got is y'=(1-(3x^2)y^3)\(3x^3y^2-1)

anonymous · Answer

That's what I get too; looks like you know what you're doing here.

anonymous · Answer

Thanks! Wait, d|dxsquareroot(xy) is ((x)(y))^1/2= 1/2(xy'+y)^-1/2?

anonymous · Answer

Close. There's chain rule and product rule going on here.
$$\large \frac{d}{dx} \sqrt{xy}=\frac{1}{2}(xy)^{-1/2} \cdot (xy'+y).$$

anonymous · Answer

Ohhh! Yeah! Thanks! :D