If u=f(x,y) where x=(e^s)(cos(t)) and y=(e^s)(sin(t)), show that (du/dx)^2 + (du/dy)^2 = (e^(-2s))((du/ds)^2+(du/dt)^2))

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If u=f(x,y) where x=(e^s)(cos(t)) and y=(e^s)(sin(t)), show that (du/dx)^2 + (du/dy)^2 = (e^(-2s))((du/ds)^2+(du/dt)^2))

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pretty complicated chain rule problem, i'd really appreciate some help on this one
what do you mean (du/dx)^2 is that the second derivative ?
no, its the square of du/dx

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no, its the square of du/dx
Its kinda confusing since usually f(x,y) is something like X^2 + y^3 something like that. Another confusing thing about the problem is that you have the derivative of u in respect to x and (du/dx) and derivative of u in respect to y (du/dy) which doesnt make sense.
i mean you can say u= x+ y then the question would make more sense but in that case you have to find (dx/dt) or (dx/ds) or vice versa with y
you mean have x+y be the function? f(x,y)=x+y?
no usually the problem is set up this way f(x,y)= z if z= w/e ( it has an x and a y) where y= (something that has s and t) and x= ( something that has s and t) but your question does not make sense to me. Are you sure you wrote it right?
yes, i copied it exactly.
what does it mean to find du/dx?
To find the derivative of the function u in respect to x for example lets say u= x^2 if you want to find du/dx = 2* x
ok well thanks anyway for your help
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YES I AM. you really are a hero! the only thing i dont understand is what du/ds and du/dt are equal to, and how you got that.
when you say u sub x times c, what is c?
Oh, sorry, I use "c" and "s" for cos(t), sin(t) here.
Where I write "NTS", that means, "need to show".
It's messy.
yes, but you have no idea how messy mine is after working on it for over an hour now. thank you so much!
du/ds and du/dt...that comes from a result in calculus (chain rule for functions of many variables).
You have u = f(x(s,t), y(s,t)). The parameters that do all the 'controlling' in the function are s and t, so you take the partial derivative with respect to them.
Is that okay?
yes, thank you.
You're welcome :)

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