Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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:

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.