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Define $F$ as $F(x,y,u,v) = x^3e^{uv} + vy^2\sin{\left(y^3\right)}$ where $u(x,y) = x^2y$ and $v(x,y) = xy^3$.

Define $f(x,y) = F(x,y,u(x,y),v(x,y)$.

Determine $\dfrac{\partial F}{\partial x}$ and $\dfrac{\partial f}{\partial x}$.

How to do this for $F$ and $f$ respectively? Aren't $F$ and $f$ the same? I'm confused. Thanks

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    What you have is $f$ is a "function" of $F$ I.e. $f(F)=F$.2012-11-07

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EDIT: Ok, your question was edited.

Try to look the "path" of the variables of your function.

By example, here we have:

$~F \to u, v, x, y~$ WHERE $~u~$ and $~v~$ depends of $~x~$ and $~y~$.

If you define a new function, $~f~$, which has $~x,~y~$ as variables, it remains only two variables (in fact).

By example, if you want to determine the partial derivative of $~f~$, just "follow" the path to "$~x~$".

$~f \to x ,~ y , ~u(x,y),~ v(x,y),~$ so you have to derivate everywhere $~x~$ is.

So, you'll have:

$$~\frac{df}{dx} + \frac{df}{du}\cdot \frac{du}{dx} + \frac{df}{dv}\cdot\frac{dv}{dx}~$$ where $~d~$ is the partial derivative.

I might be not clear as English isn't my mother-tongue, but I suggest you this link:

http://www.math.hmc.edu/calculus/tutorials/multichainrule/

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    When I said "same" it doesn't mean equals.2012-11-08