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

\[(x-a)^{3} - y ^{3} = a ^{3}\] 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.

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

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

@nuk64 , what did you get?

do u get what i did ? @nuk64

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

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

tnx @dpaInc . u r so kind

:) now i'm turning pink...

dear a is constant so it don't change

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.

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

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

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

yes

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

@dpaInc : :-) ;-)

..it's not right. :(

what's not right ?

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

plz write ur answer here , i'll check it

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

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