How do I find dy/dx for x2y+xy2=6 using implicit differentiation?

[Jim4592]

Homework Statement

find dy/dx for x²y+xy²=6

Homework Equations

The Attempt at a Solution

d/dx (x²y+xy₂) = d/dx (6)
x²*(dy/dx)+y*2x+x*2y(dy/dx)+2y=0
x²*(dy/dx)+y*2x+x*2y(dy/dx)=-2y
(dy/dx)+y*2x+x*2y(dy/dx)=(-2y/x²)
(dy/dx)+y*2x+(dy/dx)=(-2y/x²+2xy)

I'm not sure what to do now, i know that the answer is (2xy-y²/x²+2xy)

I got the denominator correct in my answer but I'm not sure how to get the numerator correctly. Did I go wrong somewhere during the process?

 

[quantumdude]

Now you algebraically solve for dy/dx.

 

[Unco]

Homework Statement

find dy/dx for x²y+xy²=6

The Attempt at a Solution

d/dx (x²y+xy²) = d/dx (6)
x²*(dy/dx)+y*2x+x*2y(dy/dx)+2y=0

Try that again. Piece by piece:
$$x^2y$$ differentiates to $$2xy + x^2y'$$
$$xy^2$$ differentiates to $$y^2 + 2xyy'$$

Then follow Tom's advice!

 

[tiny-tim]

I'm not sure what to do now, i know that the answer is (2xy-y²/x²+2xy)

I got the denominator correct in my answer but I'm not sure how to get the numerator correctly.

Hi Jim4592!

Isn't it (2xy + y²/x²+2xy)?

 

[Unco]

Isn't it (2xy + y²/x²+2xy)?

Rather,

-[/color](2xy + y²)/(x² + 2xy)

 

[tiny-tim]

oops!

thanks, Unco! ​