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Partial F/ Partial T... F= (x,y) x and y = functs of s and t...Only step 1 needed

 
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Mar21-11, 07:45 PM   #1
 

Partial F/ Partial T... F= (x,y) x and y = functs of s and t...Only step 1 needed


Wrong area sorry. Nub here! Can someone delete this.



If f(x,y) = Sqrt[x^4 + y] and x = s^2 + t^2 + 1 and y = t^2 +t*Cos(2s)

How do i find

Partial f
--------
Partial t


I made the tree diagram how f depends on x and y, which both depend on s and t.... so on my test I said it was

(Partials of all of these)


f/x*x/s*x/t + f/y*y/s*y/t


Looking back now I wonder why I put the s's in there, seems like I don't need them. (this was from a recent test and we get to redo 1 problem). My new idea is

f/x*x/t + f/x*x/y ? That seems to simple though

Can someone here point in the right direction?

Thanks
 
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Mar23-11, 06:04 AM   #2
 
[tex]\frac{\partial f}{\partial t}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial t}[/tex]
 
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3 variables, calculus 3, partial, tree diagram
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