find y'' of 2xy=y^2?

Question

I recently got my AP Calc. final exam results and I want to know how exactly to do the problems I kept getting wrong.

Answer 1

Note: product rule for 2xy to solve for y' is required.
Quotient rule for y'' is required as well.

2y + 2x*dy/dx = 2y *dy/dx
2y = 2y*dy/dx - 2x*dy/dx
2y = dy/dx(2y-2x)
2y/(2y-2x) = dy/dx

d^2y/dx^2= [(2*dy/dx)(2y-2x) - (2y)(2*dy/dx-2)]/(2y-2x)^2

But we know dy/dx=2y/(2y-2x), so we get
d^2y/dx^2= [(2*(2y/(2y-2x))(2y-2x) - (2y)(2(2y/(2y-2x))-2)]/(2y-2x)^2

d^2y/dx^2= 4y - 2y[(4y/(2y-2x)) -2] / (2y-2x)^2

NOTE: you cannot cancel the y's in the original function
2xy=y^2 (i.e.: 2x=y). THIS IS INCORRECT. You will lose answers to the solution by canceling out the y's.

Answer 2

Shay, the ANSWER IS NOT ZERO. The reason they get zero is because they divide both sides by y. REMEMBER....NEVER DIVIDE BY VARIABLES AS IT GET'S RID OF SOLUTIONS: This is the most common mistake i see by my students:

What we do is take the derivative of both sides:
2xy will be done by the chain rule
y^2 done by the power rule

2(1)(y) + 2x(y') = 2yy'
2y + 2xy' = 2yy' divide everything by 2
y + xy' = yy'
y = yy' - xy'
y = y'(y - x)
y' = y/(y-x)

Now we have to take the 2nd derivative by using the quoient rule:

y'' = (y'(y-x) - (y'-1)(y))/(y-x)^2 we know what y' is, so plug in
y'' = ((y/(y-x))(y-x) - ((y/(y-x))-1)y)/(y-x)^2
y'' = (y - ((y-y+x)/(y-x)y)/(y-x)^2
y'' = (y-(xy/(y-x)))/(y-x)^2
y" = (y(y-x) -xy)(x-y)^3
y" = (y^2 - 2xy)/(x-y)^3

Answer 3

0, they are correct

Answer 4

some interesting solutions, but the y"=0 answers are correct.

Answer 5

First, you need to complete the square, then solve for y.

1). Subtract 2xy from both sides
y^2 - 2xy = 0

2). Add x^2 to both sides
y^2 - 2xy + x^2 = x^2

3). Factor the left side to yield
(y-x)^2 = x^2

4). Take the square root of both sides
(y-x) = x

5). Add x to both sides to yield
y = 2x

To test if this is true, input this value for y into the original equation:

2x(2x) = (2x)^2
4x^2 = 4x^2

Edition:
sorry, I forgot it was asking for y'', not y...

dy/dx = 2; d^2y/dx^2 = d/dx (2) = 0

y'' = 0

Answer 6

2xy=y^2
y^2= 2xy
(y^2)/y = 2x
y = 2x
y' = 2
y'' = 0

Hope you got it :)

Answer 7

well, first off u want to get y by itself by dividing y from each side then you want to get rid of the x. i dont know the answer because im to lazy.

Answer 8

This is just y = 2x
y'' = 0

Answer 9

y^2= 2xy
y=2x
y'=2
y''=0

Answer 10

( 2 x ) ( y ) = y ²
2y + 2x (dy/dx) = 2y dy/dx
(2x - 2y) dy/dx = - 2y
(2)(x - y) dy/dx = - 2y
dy/dx = - y / (x - y)

d ² y / dx ² is given by :-
- [ (dy/dx) (x - y) - y(1 - dy/dx) ] / (x - y) ²
- [ - y - y (1 + y / (x - y) ] / (x - y) ²
- [ - 2y - y ² / (x - y) ] / (x - y) ²
- [- 2y ( x - y ) - y ² ] / (x - y)³
- [- 2xy + y ² ] / (x - y) ³
[ 2xy - y ² ] / (x - y) ³