x^3+y^3=1

y^3=1-x^3

y=1-x (take cube root of both, cube root of 1 makes 1)

now getting derivative:

y=-1

dunno if it's right, haven't done one like this =P

o_O I just noticed this when you answed, Nagao .

The question at the top of the page says find dy/dx by implicit Differentation.

I terms of finding derivative, I think you were right, but because I didn't read the question correctly, you got it wrong (inadvertently).

I thank you though: you gave me the idea that you could take the CUBE ROOT of a CUBED term. I think I can use that somewhere in this lesson, though.

You wouldn't take the cubic root for a differential.

Just use the rules of differentiation, in this case n^x = xn^(x-1), and the 1 of course becomes 0 because it's a constant.

dy/dx = y'

So for x^3 + y^3 = 1:

3x^2 + 3y^2y' = 0

3y^2y' = -3x^2

y' = -3x^2/3y^2

...and your end result:

y' = -x^2/y^2

If that looked confusing, it's (3y^2)(y') in all occurrences and (-3x^2)/(3y^2) in all occurrences.

"You wouldn't take the cubic root for a differential."

I never said this. All I said was I could take cubed root for a cubed term which means I may use it for something other than Differentiation.

"(-3x^2)/(3y^2) "

You can further simplify the expression tion

-x^2/y^2 since the 3's will cancel each other. Thanks for the help. I think I see where I went wrong.

"You wouldn't take the cubic root for a differential."

That was more of a reply to Nagao.

"You can further simplify the expression"

If you look at my reply again, you will see that I have that answer in bold under "...and your end result:"

Glad to help.

"That was more of a reply to Nagao."

Well, he only did this operation because I misworded the question.

I don't blame him, though: it kind of looks like an algebraic problem.