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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