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

Question

anonymous · Answer

$$(x-a)^{3} - y ^{3} = a ^{3}$$
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.

anonymous · Answer

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

anonymous · Answer

$$3(x-a)^2 -3y'y^2=0$$

anonymous · Answer

@nuk64 , what did you get?

anonymous · Answer

do u get what i did ? @nuk64

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

tnx @dpaInc  . u r so kind

anonymous · Answer

:)  now i'm turning pink...

anonymous · Answer

dear a is constant so it don't change

anonymous · Answer

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.

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

yes

anonymous · Answer

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

anonymous · Answer

@dpaInc  : :-) ;-)

anonymous · Answer

..it's not right. :(

anonymous · Answer

what's not right ?

anonymous · Answer

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

anonymous · Answer

plz write ur answer here , i'll check it

anonymous · Answer

kay, well i got this:
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