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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

Mathematics
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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?
use the quotient rule
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 ?

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Other answers:

You should try that again. The whole numerator is a mess.
ok..wait
someby pls help me..i dont get it..... :{
Numerator: (x+2y)(-y') - (-y)(1+2y') Look at it very carefully.
oh yeahh my numerator is really a mess..then what should i do ?
what should i do ~~~~~ omgg
even i tried this more than once, and i don't get numerator = 8, is anything else given ?
the original question .
idk what im doing -.-
first question
ohhh...so, xy+y^2 =4.... then its easy, did you get that numerator first ? next thing will be to simplify it.
got this : (x+2y)(-y') - (-y)(1+2y') first ?
yes i got it just now .haha
then ?
now use, y'=-y/(x+2y)
yes im using that..then quotient rule ?
(x+2y)(-y') - (-y)(1+2y') is the result of numerator AFTER applying quotient rule, now you have to simplify
Just for the record, it is ALWAYS more beneficial to show the ORIGINAL problem statement.
yeah, that would have saved me few minutes...
to be honest im stucking again
what u got after substituting y'=-y/(x+2y) in that numerator ?
\[2y-\frac{ 2y^2 }{ x+2y }\] what about this ? for that numerator
\[\frac{ 2xy+2y }{ (x+2y)^3 }\]
this ???????
then subtitute (0,2) into that equation ?
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
\(2y-\frac{ 2y^2 }{ x+2y }\) was correct
\[\frac{ 2xy+2y^2}{(x+2y)^3}\]
this ????am i right ??
yeah, factoring out 2 will give you 2 (xy+y^2) and xy+y^2 as per the question is 4
ohhhh........i see ur answer........and i totally got it ! thanks a lot @hartnn !!:) really appreciate ^_^
welcome ^_^

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