Question

Find the second derivative of xy +y^2 = 1.
The first derivative is -y(x + 2y)^(-1).
The text book answer is 2(x + 2y)^-3, but I've been unable to get that answer.

Answer 1

y+xy'+2yy'=0
y'=-y/(x+2y)

Answer 2

Dx(xy +y^2 = 1)

Dx(xy) +Dx(y^2) = Dx(1)

(x'y + xy') +(2y y') = (0)

x'y + y'(x +2y) = 0

y'(x +2y) = -x'y

y' = -x'y/(x +2y)

Answer 3

Thanks, but I have the first derivative. I need the second.

Answer 4

since dx/dx = 1 .. x'=1

Dx[y' = -y/(x +2y)]

Dx[y'] = Dx[-y (x +2y)^-1]

y'' = -y'(x+2y^-1) + -y -(x+2y)^-2 Dx(x+2y)

Answer 5

Dx(x+2y)

Dx(x) + Dx(2y)

x' + 2 y'

1 + 2y'

\[y'' = -y'(x+2y^{-1}) + y(x+2y)^{-2} (1+2y')\]
\[y'' = \frac{-y'}{(x+2y)^{-1}} + \frac{y (1+2y')}{(x+2y)^{-2}}\]
\[y'' = \frac{-y'}{(x+2y)^{-1}} + \frac{y+2yy'}{(x+2y)^{-2}}\]

Answer 6

get like denoms and it should simplify to the text

Answer 7

but I think I see a slight error in my math to text skills lol

Answer 8

denominator (x+2y)?

Answer 9

also(x+2y)2

Answer 10

you know; Newton was wrong alot too lol

Answer 11

the 1st derive is good; the second needs som etender lovin care applied to finesse it to the proper look

Answer 12

the prob is just with denominator

Answer 13

(x+2y)^2 is the lcm

Answer 14

\[Dx[y' = -\frac{y}{(x +2y)}]\]
\[y'' = -\frac{(x+2y)y' - y(1+2y')}{(x +2y)^2}\]

Answer 15

y''=(-xy'+y)/(x+2y)^2

Answer 16

y''=xy/(x+2y)+y
------------
(x+2y)^2

Answer 17

even then it doesn't match the solution:S
all becoz of typing :)

Answer 18

\[y'' = -\frac{y'x +2yy' -y-2yy'}{(x +2y)^2}\]
\[y'' = -\frac{y'x -y}{(x +2y)^2}\]
\[y'' = -\frac{(-\frac{y}{(x +2y)})x -y}{(x +2y)^2}\]

Answer 19

=2xy+2y^2/(x+2y)^3

Answer 20

\[y'' = -\frac{-\frac{yx}{(x +2y)} -\frac{y(x+2y)}{(x+2y)}}{(x +2y)^2}\]
\[y'' = -\frac{-yx - y(x+2y)} {(x+2y)(x +2y)^2}\]
\[y'' = -\frac{-yx - yx-2y^2} {(x +2y)^3}\]

Answer 21

2(xy+y^2)/(x+2y)^3=2(x+2y)^-3

Answer 22

yayyy...got it :)

Answer 23

xy+y^2=1

Answer 24

\[y'' = -\frac{-2yx-2y^2} {(x +2y)^3}\]
\[y'' = \frac{2y(x+y)} {(x +2y)^3}\]

Answer 25

newton in wrong direction again ;)

Answer 26

lol

Answer 27

taking 2 outside

Answer 28

go on, m so bad in latex

Answer 29

Here's one of my attempts
y'' = [y(1 + 2 y' - (x +2y)]/(x+2y)^2
y + 2y(-y/(x+2y) - (x+2y)/ (x +2y)^2
= y (x+2y) -2y^2 - (x+2y)^2]/(x +2y)^3
=then I'm having trouble

Answer 30

Thanks to both of you for hanging in there.