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

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