Proof Involving Partial Derivatives Chain Rule

**runningeagle** · January 25th 2010, 12:34 PM

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).

Thank your for your help.

**Opalg** · January 26th 2010, 12:03 AM

Originally Posted by runningeagle

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

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