Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

@nuk64 , what did you get?

Answer 5

do u get what i did ? @nuk64

Answer 6

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

Answer 7

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

Answer 8

tnx @dpaInc . u r so kind

Answer 9

:) now i'm turning pink...

Answer 10

dear a is constant so it don't change

Answer 11

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.

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

yes

Answer 16

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

Answer 17

@dpaInc : :-) ;-)

Answer 18

..it's not right. :(

Answer 19

what's not right ?

Answer 20

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

Answer 21

plz write ur answer here , i'll check it

Answer 22

kay, well i got this: