Question

second derivtive of xy+y^2=1. got a complicated ans. please help with this.

Answer 1

xy+y^2=1

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

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

Answer 2

(xy+2y+x)/(2y+3)^3 i got this as 2nd derivative

Answer 3

since x' = 1; and x'' = 0


y' + y' + xy'' +2 y' + 2y y'' = 0

4y' + y''(x + 2y) = 0

y'' = -4y'/(x + 2y)

Answer 4

need to find d^2y/dx^2
first find dy/dx
xy'+y+2yy'=0
now differentiate again
2y'+xy''+2y'^2+2yy''=0
we get y''(x+2y)=-(2y'+2y'^2)
y''=-(2y'+2y'^2)/(x+2y)

Answer 5

dealing with the y'

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

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

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

y'' = -4y'/(x + 2y)

y'' = -4(-y/x+2y)/(x + 2y)

y'' = 4y/(x + 2y)^3

is what i get, but id have to dbl check

Answer 6

1st derivative is the same as yours

Answer 7

\[y+xy'+2yy'=0\]
\[(x+2y)y'=-y\]
\[y'=\frac{y}{2x+y}\] is that i get, then one more time

Answer 8

oops i am off by a minus sign at the end

Answer 9

for2nd derivative i used quotient rule

Answer 10

yes, and then you will have y' in your answer, so you should substitute back
\[y'=\frac{-y}{x+2y}\] and do some more algebra. big fun

Answer 11

no fun at all. i think i messed up in algebra

Answer 12

it looks like amistre did it already. i am late as usual

Answer 13

ugh

Answer 14

:) but i reserve the right to be wrong lol

Answer 15

ok i get
\[\frac{y-xy'}{(x+2y)^2}\] as a first step. look good?

Answer 16

now i'm getting (xy+y^2+y^2+xy)/(2y+3)^3. replacing the xy+y^2 with 1 frm question.
2/(2y+3)^3

Answer 17

is that ok

Answer 18

could be but it is not what i got. i am fairly sure of this line
\[\frac{y-xy'}{(x+2y)^2}\]

Answer 19

mines off a bit ... bummer

Answer 20

oh once again i am off by a minus sign. let me check again

Answer 21

ok i got what i think is right
\[\frac{y+\frac{xy}{x+2y}}{(x+2y)^2}=\frac{y(x+2y)+xy}{(x+2y)^3}=\frac{xy+2y^2+xy}{(x+2y)^3}=\frac{1+xy}{(x+2y)^3}\]

Answer 22

yep that's what i got as well. thanx