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where is v in your formula?

F(u(s,t),v(s,t)). where... yes. it is

W(1,0)=F(u(1,0),v(1,0))and we are told that u(1,0)=2 and v=(1,0)=3, so
W(1,0)=F(2,3)

looks like multivariable chain rule problem ...

@experimentX yes, it is.
@TuringTest and then?

\[\frac{ \delta F }{ \delta u }= F(2,3)\]

is it right?

the second formula you wrote is not right, the first is

\[ F_u(u(1,0), v(1,0))\]

you can't simply equate the partial differential with the function itself.

ok, sorry, I think tooooo far. waste your time

its okay ... the partial with respect to time will give you velocity along x and y axis

i've been doing this for last 20 years

and never ask yourself the silly question like me?

sure ... anytime!!

Bye bye.