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

  • Thread starter gillgill
  • Start date
  • #1
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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?
 

Answers and Replies

  • #2
2,209
1
Both are, but the second is alot sloppier.
 
  • #3
128
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another question:
x^3+y^3=6xy

3x^2+(3y^3)y'=...?
how do u differentiate 6xy?
 
  • #5
Zurtex
Science Advisor
Homework Helper
1,120
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gillgill said:
how do u differentiate 6xy?
Product rule! :wink:
 
  • #6
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i only learnt how to differentiate using prodct rule with two terms....is it the same with 3 terms?
 
  • #7
2,209
1
6 is a constant, dont even worry about it.
 
  • #8
128
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so....(6xy')+(6y)....?
 
  • #9
2,209
1
6xy

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

(u(x)v(x))' = u'(x)v(x) + v'(x)u(x)
 
  • #10
128
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did i start this right?
x^2+y^2+2x-4y-20=0
2x+(2y)(y')+2+4y'=0
 
  • #11
998
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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.
 
  • #12
128
0
o..i see it...thanks
 

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