Proof Involving Partial Derivatives Chain Rule

[runningeagle]

Hello,

"z=f(x,y)

x=e[sup:amgdgwgt]s[/sup:amgdgwgt]cos(t)

y=e[sup:amgdgwgt]s[/sup:amgdgwgt]sin(t)

show d[sup:amgdgwgt]2[/sup:amgdgwgt]z/dx[sup:amgdgwgt]2[/sup:amgdgwgt]+d[sup:amgdgwgt]2[/sup:amgdgwgt]z/dy[sup:amgdgwgt]2[/sup:amgdgwgt] = e^(-2s)[d[sup:amgdgwgt]2[/sup:amgdgwgt]z/ds[sup:amgdgwgt]2[/sup:amgdgwgt]+ d[sup:amgdgwgt]2[/sup:amgdgwgt]/dt[sup:amgdgwgt]2[/sup:amgdgwgt]"

I know how to do the chain rule. I can find dx/ds dx/dt dy/ds and dy/dt but I don't know how to find anything involving z because z is only defined as f(x,y).

Thank you for your help

 

[Deleted member 4993]

Hello,

"z=f(x,y)

x=e[sup:1krf8fx7]s[/sup:1krf8fx7]cos(t)

y=e[sup:1krf8fx7]s[/sup:1krf8fx7]sin(t)

show d[sup:1krf8fx7]2[/sup:1krf8fx7]z/dx[sup:1krf8fx7]2[/sup:1krf8fx7]+d[sup:1krf8fx7]2[/sup:1krf8fx7]z/dy[sup:1krf8fx7]2[/sup:1krf8fx7] = e^(-2s)[d[sup:1krf8fx7]2[/sup:1krf8fx7]z/ds[sup:1krf8fx7]2[/sup:1krf8fx7]+ d[sup:1krf8fx7]2[/sup:1krf8fx7]/dt[sup:1krf8fx7]2[/sup:1krf8fx7]"

I know how to do the chain rule. I can find dx/ds dx/dt dy/ds and dy/dt but I don't know how to find anything involving z because z is only defined as f(x,y).

Thank you for your help

dz/dx = dz/ds * ds/dx + dz/dt * dt/dx