What Is the Correct Approach to Solve These Implicit Differentiation Problems?

[pr0blumz]

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?

 

[rock.freak667]

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

Where'd you get that extra minus sign from when differentiating -10xy?

 

[pr0blumz]

I differentiated (-10xy)

 

[rock.freak667]

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

That -ve sign should be a + sign.

 

[pr0blumz]

I'm still getting (5y-8x)/(3y-5x)

 

[rock.freak667]

I'm still getting (5y-8x)/(3y-5x)

What answer are you supposed to get?

 

[pr0blumz]

(8x-5y)/(5x-3y)

 

[rock.freak667]

(8x-5y)/(5x-3y)

Well multiply both the numerator and denominator by -1.

 

[pr0blumz]

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!

 

[HallsofIvy]

As rockfreak667 said before, you already had the correct answer:
(5y-8x)/(3y-5x)= (8x-5y)/(5x-3y)

 

[pr0blumz]

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?

 

[rock.freak667]

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 2x^1/2y^1/2