# Thread: Proof Involving Partial Derivatives Chain Rule

1. ## Proof Involving Partial Derivatives Chain Rule

Hi,

"if u=f(x,y) is a differentiable function with x=e^s((cos(t)) y=e^s((sin(t)) show that show d2u/dx2+d2u/dy2 = e^(-2s)[d2u/ds2+ d2u/dt2

I know how to find dx/dt dy/ds etc, but I don't know how to find anything involving u because it is only defined as f(x,y).

Use the chain rule: $\displaystyle \frac{\partial u}{\partial s} = \frac{\partial u}{\partial x}\frac{\partial x}{\partial s} + \frac{\partial u}{\partial y}\frac{\partial y}{\partial s}$. Then when you differentiate a second time you have to use the product rule as well as the chain rule, for example $\displaystyle \frac{\partial }{\partial s}\Bigl(\frac{\partial u}{\partial x}\frac{\partial x}{\partial s}\Bigr) = \Bigl(\frac{\partial^2 u}{\partial x^2}\frac{\partial x}{\partial s} + \frac{\partial^2 u}{\partial y\partial x}\frac{\partial y}{\partial s}\Bigr)\frac{\partial x}{\partial s} + \frac{\partial u}{\partial x}\frac{\partial^2 x}{\partial s^2}$. That way, you can find an expression for $\displaystyle \frac{\partial^2 u}{\partial s^2} + \frac{\partial^2 u}{\partial t^2}$.