How to differentiate 2x/(x+y) = y, implicitly?

[gillgill]

differentiate
2x/(x+y)=y

Method1:
2x=xy+y^2
d/dx(2x)=d/dx(xy+y^2)
2=y+xy'+2y(y')
y'=(2-y)/(x+2y)

Method2:
2x/(x + y) = y
y' = [2(x + y) - (2x)(1 + y')]/(x + y)^2
[(x + y)^2 + 2x]y' = 2(x + y) - 2x
y' = 2y/[(x + y)^2 + 2x]

Which method is using "implicit" differentiation?

 

[whozum]

Both are, but the second is a lot sloppier.

 

[gillgill]

another question:
x^3+y^3=6xy

3x^2+(3y^3)y'=...?
how do u differentiate 6xy?

 

[UrbanXrisis]

product rule

 

[Zurtex]

how do u differentiate 6xy?

Product rule!

 

[gillgill]

i only learned how to differentiate using prodct rule with two terms...is it the same with 3 terms?

 

[whozum]

6 is a constant, don't even worry about it.

 

[gillgill]

so...(6xy')+(6y)...?

 

[whozum]

6xy

u(x) = 6x, v(x) = y

(u(x)v(x))' = u'(x)v(x) + v'(x)u(x)

 

[gillgill]

did i start this right?
x^2+y^2+2x-4y-20=0
2x+(2y)(y')+2+4y'=0

 

[Data]

did i start this right?
x^2+y^2+2x-4y-20=0
2x+(2y)(y')+2+4y'=0

you seem to have made a sign error, but other than that it's okay.

 

[gillgill]

o..i see it...thanks