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gillgill
Apr23-05, 06:36 PM
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
Apr23-05, 06:42 PM
Both are, but the second is alot sloppier.

gillgill
Apr23-05, 07:55 PM
another question:
x^3+y^3=6xy

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

UrbanXrisis
Apr23-05, 07:59 PM
product rule

Zurtex
Apr23-05, 08:00 PM
how do u differentiate 6xy?
Product rule! :wink:

gillgill
Apr23-05, 08:19 PM
i only learnt how to differentiate using prodct rule with two terms....is it the same with 3 terms?

whozum
Apr23-05, 08:20 PM
6 is a constant, dont even worry about it.

gillgill
Apr23-05, 08:28 PM
so....(6xy')+(6y)....?

whozum
Apr23-05, 08:54 PM
6xy

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

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

gillgill
Apr23-05, 11:37 PM
did i start this right?
x^2+y^2+2x-4y-20=0
2x+(2y)(y')+2+4y'=0

Data
Apr23-05, 11:40 PM
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
Apr23-05, 11:43 PM
o..i see it...thanks