Question

I have the answer can somone please check me ....
Find (dy/dx)
x^3+y^3=3x

Answer 1

\(y^3=3x-x^3\)
\(y=\sqrt[3]{3x-x^3}\)
\(y'=-\frac{x-3x^2}{3\sqrt[3]{(3x-x^3)^2}}\)

Answer 2

um I am really confused that is not what I got

Answer 3

it is just chain rule.

Answer 4

you could write it also like this: \(y=(3x-x^3)^{\frac{1}{3}}\)
and diferentiate it

Answer 5

\[\frac{ 1-x^2 }{ y^2}\]Yes but look at the question again the answer that I got is

Answer 6

no idea how you got that

Answer 7

well you have to find\[\frac{ dy }{ dx }\] of the x^3+y^3=3x

Answer 8

so first express y as function of x

Answer 9

ok got that

Answer 10

or use implicit differentiation

Answer 11

what is the final answer that you get for the overall problem

Answer 12

I already typed it in my 1º answer

Answer 13

can you please retype what you think the answer is again please

Answer 14

is the last line of my first answer

Answer 15

ok but all my multiple choice answers I don't have that to choose from any of them

Answer 16

what are your choices

Answer 17

one is 3+3x^2+3y^2

Answer 18

another is \[\frac{ 1-3x^2 }{ y^2 }\]

Answer 19

ups I made a typo there

Answer 20

where

Answer 21

can you retype your answer please.

Answer 22

in the numerator should be 3-3x^2

Answer 23

ok well I still don't have that for any of my choices

Answer 24

another answer is 3x^2+3y^2

Answer 25

another one is \[\frac{ x^2 }{ y^2}\]

Answer 26

another choice is \[\frac{ 3-x^2 }{ y^2}\]

Answer 27

another chice is \[\frac{ -x^2 }{ y^2 }\]

Answer 28

you could also write it as:
\(\huge \frac{1-x^2}{y^2}\)

Answer 29

another choice is 3-3x^2-3y^2

Answer 30

ok thanks I gave that choice for my first answer but you said it was not right

Answer 31

my bad

Answer 32

yeah

Answer 33

its ok thanks for your help:)

Answer 34

yw