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OpenStudy (anonymous):

Find dy/dx in terms of x and y if (x-a)^3 - y^3 = a^3.

OpenStudy (anonymous):

\[(x-a)^{3} - y ^{3} = a ^{3}\] i typed it out better - this is what i meant. i need to differentiate that, and i've tried it 3 times, but i can't seem to get it.

OpenStudy (anonymous):

@nuk64 : the easiest way is to understand y changes with x ! so

OpenStudy (anonymous):

\[3(x-a)^2 -3y'y^2=0\]

OpenStudy (anonymous):

@nuk64 , what did you get?

OpenStudy (anonymous):

do u get what i did ? @nuk64

OpenStudy (anonymous):

just solve for y' in Roya's answer...

OpenStudy (anonymous):

something really long and complicated >< cuz i expanded (x-a)^3 and then tried differentiating. but the 'a's didn't cancel

OpenStudy (anonymous):

tnx @dpaInc . u r so kind

OpenStudy (anonymous):

:) now i'm turning pink...

OpenStudy (anonymous):

dear a is constant so it don't change

OpenStudy (anonymous):

yeah but when you expand it, and then have a alone, it goes to zero, right? cuz it's a constant? but when it's in the middle, then you can't get rid of it.. i think.

OpenStudy (anonymous):

i mean u cant say da/dx and for (x-a)^3 suppose x-a=u dx=du then use chain way

OpenStudy (anonymous):

let me expend it too ! (x-a)3=x^3-3ax^2+3a^2x-a^3 am i right ?

OpenStudy (anonymous):

oh okay. so then the answer's just: \[(x-a)^{2}\div y ^{2}\]?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

and when u extend it u will have 3x^2-6ax +3a^2 now just factor 3 !

OpenStudy (anonymous):

@dpaInc : :-) ;-)

OpenStudy (anonymous):

..it's not right. :(

OpenStudy (anonymous):

what's not right ?

OpenStudy (anonymous):

the answer i got :O where you said to just solve for y-prime. i did that. but yeah, it didn't work. :/

OpenStudy (anonymous):

plz write ur answer here , i'll check it

OpenStudy (anonymous):

kay, well i got this: |dw:1338186485549:dw|

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