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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
 one year ago
 one year ago
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
 one year ago
 one year ago

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tkhunnyBest ResponseYou've already chosen the best response.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?
 one year ago

Ika_KimBest ResponseYou've already chosen the best response.1
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 ?
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.1
You should try that again. The whole numerator is a mess.
 one year ago

Ika_KimBest ResponseYou've already chosen the best response.1
someby pls help me..i dont get it..... :{
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.1
Numerator: (x+2y)(y')  (y)(1+2y') Look at it very carefully.
 one year ago

Ika_KimBest ResponseYou've already chosen the best response.1
oh yeahh my numerator is really a mess..then what should i do ?
 one year ago

Ika_KimBest ResponseYou've already chosen the best response.1
what should i do ~~~~~ omgg
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
even i tried this more than once, and i don't get numerator = 8, is anything else given ?
 one year ago

Ika_KimBest ResponseYou've already chosen the best response.1
the original question .
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
ohhh...so, xy+y^2 =4.... then its easy, did you get that numerator first ? next thing will be to simplify it.
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
got this : (x+2y)(y')  (y)(1+2y') first ?
 one year ago

Ika_KimBest ResponseYou've already chosen the best response.1
yes i got it just now .haha
 one year ago

Ika_KimBest ResponseYou've already chosen the best response.1
yes im using that..then quotient rule ?
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
(x+2y)(y')  (y)(1+2y') is the result of numerator AFTER applying quotient rule, now you have to simplify
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.1
Just for the record, it is ALWAYS more beneficial to show the ORIGINAL problem statement.
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
yeah, that would have saved me few minutes...
 one year ago

Ika_KimBest ResponseYou've already chosen the best response.1
to be honest im stucking again
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
what u got after substituting y'=y/(x+2y) in that numerator ?
 one year ago

Ika_KimBest ResponseYou've already chosen the best response.1
\[2y\frac{ 2y^2 }{ x+2y }\] what about this ? for that numerator
 one year ago

Ika_KimBest ResponseYou've already chosen the best response.1
\[\frac{ 2xy+2y }{ (x+2y)^3 }\]
 one year ago

Ika_KimBest ResponseYou've already chosen the best response.1
then subtitute (0,2) into that equation ?
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
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
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
\(2y\frac{ 2y^2 }{ x+2y }\) was correct
 one year ago

Ika_KimBest ResponseYou've already chosen the best response.1
\[\frac{ 2xy+2y^2}{(x+2y)^3}\]
 one year ago

Ika_KimBest ResponseYou've already chosen the best response.1
this ????am i right ??
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
yeah, factoring out 2 will give you 2 (xy+y^2) and xy+y^2 as per the question is 4
 one year ago

Ika_KimBest ResponseYou've already chosen the best response.1
ohhhh........i see ur answer........and i totally got it ! thanks a lot @hartnn !!:) really appreciate ^_^
 one year ago
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