# implicit functions

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 are looking for is the multivariable chain rule. – Daryl Nov 6 '12 at 4:20
chain rule tutorials doesn't have this situation. – Frank Xu Nov 7 '12 at 0:49
What you have is $f$ is a "function" of $F$ I.e. $f(F)=F$. – Daryl Nov 7 '12 at 1:15

EDIT: Ok, your question was edited.

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

By example, here we have:

F -> 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 -> x , y , u(x,y), v(x,y), so you have to derivate everywhere x is.

So, you'll have:

df/dx + (df/du)(du/dx) + (df/dv)(dv/dx) where d is the partial derivative.

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

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

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is df/dx + (df/du)(du/dx) + (df/dv)(dv/dx) for f? how about F – Frank Xu Nov 6 '12 at 5:15
for me it's the same – Provost Nov 6 '12 at 5:28
no , you're wrong – Frank Xu Nov 7 '12 at 0:19
@Frank Xu If you think this solution is wrong, can you describe why you think it is wrong? – Daryl Nov 7 '12 at 1:18
When I said "same" it doesn't mean equals. – Provost Nov 8 '12 at 17:22