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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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    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.2011-10-24

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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 $\frac{3\sqrt{u}}{2}+3y^2\frac{dy}{du}=0$

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    @GerryMyerson: implicit differentiation comes in the next paragraph. But you are right, that seems to be OP writing, not the teacher. Maybe my first two sentences are all that was asked?2011-10-24