Partial derivitives chain rule proof

[ProPatto16]

Homework Statement

If u=f(x,y) where x=e^scost and y=e^ssint

show that d²u/dx²+d²u/dy² = e^-2s[d²u/ds²+d²u/dt²

The Attempt at a Solution

i have no idea!

question though, do the partial derivitives have to be solved and expanded then just show that one side equals the other or can it be proven by manipulating the terms as they are?

the only relevant chain rule is this one

du/ds=du/dx*dx/ds+du/dy*dy/ds and du/dt=du/dx*dx/dt+du/dy*dy/dt

this gives the first step of the partial derivitives to some of the question. how would i go about finding the second partial derivative?

any help would be great!

 

[HallsofIvy]

You find the second partial derivative by doing it again!

For example, $x= e^s cos(t)$ and $y= e^s sin(t)$ so
$$\frac{\partial f}{\partial s}= \frac{\partial f}{\partial x}(e^s cos(t)+ \frac{\partial f}{\partial y}(e^s sin(t)$$

Then
$$\frac{\partial^2 f}{\partial s^2}= \frac{\partial}{\partial s}\left(\frac{\partial f}{\partial x}(e^s cos(t)+ \frac{\partial f}{\partial y}(e^s sin(t)\right)$$
$$= \frac{\partial}{\partial s}\left(\frac{\partial f}{\partial x}\right)(e^s cos(t))+$$$$\frac{\partial f}{\partial x}\left(\frac{\partial}{\partial s}\right)\left(\frac{\partial}{\partial s}\right)\left(e^s cos(t)\right)+$$$$\frac{\partial f}{\partial y}(e^s sin(t))+ \frac{\partial}{\partial s}\left(e^s sin(t)\right)$$
$$= \frac{\partial^2 f}{\partial x^2}(e^s cos(t))^2+\frac{\partial f}{\partial x}(e^s cos(t)) \frac{\partial^2 f}{\partial x^2}(e^s sin(t))^2+ \frac{\partial f}{\partial x}(e^s sin(t))$$

 

[ProPatto16]

in the last line... should there be a + in the middle there?

 

[ProPatto16]

assuming there is meant to be a plus in there... and continuing on...

that last line simplifies to:

= d²f/dx²(e^2scos²t+e^2ssin²t) + df/dx(e^scost+e^ssint)

= d²f/dx²[e^2s(cos²t+sin²t)] + df/dx(e^scost+e^ssint)

now i can see that cos²t+sin²t = 1 so first term becomes d²f/dx²[e^2s]

then when you did it with d²f/dy² you would get d²f/dy²[e^2s] for the first term then a similar second second, then i can see how the proof would work out


but i can't get rid of the second terms df/dx(e^scost+e^ssint) and the df/dy one ?

 

[ProPatto16]

hold up...

f in terms of x and y would just be f=x+y right? without making the substitutions for the equations of x and y... therefore df/dx would be zero? then it all works out!