Question

ive tried to prove it many times but , i still didnt get it.... if dy/dx = -y/x+2y prove that d''y/dx''=8/(x+2y)^3

Answer 1

Do you mean \(\dfrac{-y}{x+2y}\)? That is NOT what you have written. Folks in calculus should know about the Order of Operations.

Have you considered the Quotient Rule for differentiation?

Answer 2

use the quotient rule

Answer 3

really ? sorry i dont know how to write it in the right way...yes i use that quotient rule , but then i got this\[\frac{ -\frac{ dy }{ dx }.x+2y-(2y.\frac{ dy }{ dx }) }{ (x+2y)^2 }\] then what should i do ?

Answer 4

You should try that again. The whole numerator is a mess.

Answer 5

ok..wait

Answer 6

someby pls help me..i dont get it..... :{

Answer 7

Numerator: (x+2y)(-y') - (-y)(1+2y')

Look at it very carefully.

Answer 8

oh yeahh my numerator is really a mess..then what should i do ?

Answer 9

what should i do ~~~~~ omgg

Answer 10

even i tried this more than once, and i don't get numerator = 8,
is anything else given ?

Answer 11

the original question .

Answer 12

idk what im doing -.-

Answer 13

first question

Answer 14

ohhh...so, xy+y^2 =4.... then its easy,
did you get that numerator first ?
next thing will be to simplify it.

Answer 15

got this : (x+2y)(-y') - (-y)(1+2y') first ?

Answer 16

yes i got it just now .haha

Answer 17

then ?

Answer 18

now use, y'=-y/(x+2y)

Answer 19

yes im using that..then quotient rule ?

Answer 20

(x+2y)(-y') - (-y)(1+2y') is the result of numerator AFTER applying quotient rule, now you have to simplify

Answer 21

Just for the record, it is ALWAYS more beneficial to show the ORIGINAL problem statement.

Answer 22

yeah, that would have saved me few minutes...

Answer 23

to be honest im stucking again

Answer 24

what u got after substituting y'=-y/(x+2y)
in that numerator ?

Answer 25

\[2y-\frac{ 2y^2 }{ x+2y }\] what about this ? for that numerator

Answer 26

\[\frac{ 2xy+2y }{ (x+2y)^3 }\]

Answer 27

this ???????

Answer 28

then subtitute (0,2) into that equation ?

Answer 29

its actually this :
\(\huge \frac{ 2xy+2y^2 }{ (x+2y)^3 }=\frac{ 2(xy+y^2) }{ (x+2y)^3 }\)
now use xy+y^2 =4
and you are done

Answer 30

\(2y-\frac{ 2y^2 }{ x+2y }\) was correct

Answer 31

\[\frac{ 2xy+2y^2}{(x+2y)^3}\]

Answer 32

this ????am i right ??

Answer 33

yeah, factoring out 2 will give you 2 (xy+y^2)
and xy+y^2 as per the question is 4

Answer 34

ohhhh........i see ur answer........and i totally got it ! thanks a lot @hartnn !!:) really appreciate ^_^

Answer 35

welcome ^_^