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Implicit Differentiation

  • Thread starter pr0blumz
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  • #1
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Homework Statement


8x^2-10xy+3y^2=26


2. The attempt at a solution

(8)(2x)-(-10x)y'+(y)(-10)+(3)(2y)y'=0

16x+10x(y')-10y+6y(y')=0

y'(10x+6y)+16x-10y=0

y'(10x+6y)=10y-16x

y'=(10y-16x)/(10x+6y)

y'=(5y-8x)/(5x+3y)

I know I'm doing something wrong but I can't see it for myself. Can someome point me into the right direction?
 

Answers and Replies

  • #2
rock.freak667
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(8)(2x)-(-10x)y'+(y)(-10)+(3)(2y)y'=0
Where'd you get that extra minus sign from when differentiating -10xy?
 
  • #3
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I differentiated (-10xy)
 
  • #4
rock.freak667
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(8)(2x)-(-10x)y'+(y)(-10)+(3)(2y)y'=0
That -ve sign should be a + sign.
 
  • #5
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I'm still getting (5y-8x)/(3y-5x)
 
  • #6
rock.freak667
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I'm still getting (5y-8x)/(3y-5x)
What answer are you supposed to get?
 
  • #7
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(8x-5y)/(5x-3y)
 
  • #8
rock.freak667
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(8x-5y)/(5x-3y)
Well multiply both the numerator and denominator by -1.
 
  • #9
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Only thing that I see happened was that y' was moved to the right side of the equation therefore changing the signs. After I done it that way, I got the correct answer. Go figure!
 
  • #10
HallsofIvy
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As rockfreak667 said before, you already had the correct answer:
(5y-8x)/(3y-5x)= (8x-5y)/(5x-3y)
 
  • #11
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As rockfreak667 said before, you already had the correct answer:
(5y-8x)/(3y-5x)= (8x-5y)/(5x-3y)
I have another problem as well.

x + (sqrtx)(sqrty) = 2y

1 + (x^1/2)/(2y^1/2)y' + (y^1/2)/(2x^1/2) = 2y'

My question is why I suppose to multiply both sides by (2x^1/2 * y^1/2) and not both of the denominators?
 
  • #12
rock.freak667
Homework Helper
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I have another problem as well.

x + (sqrtx)(sqrty) = 2y

1 + (x^1/2)/(2y^1/2)y' + (y^1/2)/(2x^1/2) = 2y'

My question is why I suppose to multiply both sides by (2x^1/2 * y^1/2) and not both of the denominators?
the common denominator of those the terms (not the 1 and not the 2y') is 2x1/2y1/2
 

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