# Could someone explain getting a derivative when you have $y=f(x^2)$ added on

So, if given a problem $y^3 + x^3 = 27$, and the statement $y = f(x^2)$,

Could someone explain what it means to find $y'$? (using implicit differentiation) What does the $y = f(x^2)$ mean? Would this be equivalent to finding $dy/dx^2$?

I'm just completely confused about this (and the usage of the chain rule... where how $v(x) = 1$) and would love some help. Sorry if this is too broad a question.

This isn't a textbook problem - our teacher wrote the two statements on the board and told us to "solve." - so I don't know the "exact" problem - is that problematic? Am I expected to be able to solve it like this?

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That's a very specific question... but not too specific :) –  The Chaz 2.0 Oct 24 '11 at 19:24
Can you quote the problem exactly, please? –  Arturo Magidin Oct 24 '11 at 19:27
If you don't know the exact problem, it can be hard to give the desired answer. –  Ross Millikan Oct 24 '11 at 21:03
I've been teaching long enough to know that what a student says a teacher said and what the teacher actually said can be two very different things. So, I'm skeptical about your story. But, giving you the benefit of the doubt, if your teacher really presented the problem this way, then your teacher was having a very bad day, because, as presented, this "problem" makes no sense. –  Gerry Myerson Oct 24 '11 at 21:30

If we solve the equation for $y$, we get $y=\sqrt[3]{27-x^3}$, but you are supposed to write it as a function of $x^2$, which we can call $u$. $y=\sqrt[3]{27-u^{(\frac{3}{2})}}$, which works fine for $x\ge0$. Presumably then you are asked to find $\frac{dy}{du}$. This is a reasonable chain rule problem.
To do this implicitly, start with $x^3+y^3=27$, write it as $u^{(\frac{3}{2})}+y^3=27$and take the derivative with respect to $u$ of each side. You get $\frac{3\sqrt{u}}{2}+3y^2\frac{dy}{du}=0$
According to OP, the teacher wrote $y^3+x^3=27$, and $y=f(x^2)$, and said "solve". I see no mention of derivatives, implicit or otherwise, in what OP said the teacher said. Still, you may have read correctly between the lines, this may be what the teacher wanted. My own guess is that the teacher was asking two entirely separate questions: find $dy/dx$ if $y^3+x^3=27$, and find $dy/dx$ if $y=f(x^2)$. –  Gerry Myerson Oct 24 '11 at 21:35