Multi-variable chain rule

**LLemon** · April 25th 2010, 12:54 AM

Question

Suppose z=f(x-ct) + g(x+ct), where c is a constant. Show that

z(tt)=c^(2)z(xx)

Where z(tt) means the second partial derivative of z with respect to t, as for z(xx) as well.

I know this involves some application of the chainn rule but I can't seem to prove it at all... please help!

**Random Variable** · April 25th 2010, 01:49 AM

$\frac{\partial^{2} }{\partial x^{2}} z (x,t) = f''(x-ct) + g''(x+ct)$

$\frac{\partial}{\partial t} z(x,t) = -cf'(x-ct) + cg'(x+ct)$ (chain rule)

$\frac{\partial^{2} }{\partial t^{2}} z(x,t)= c^{2}f''(x-ct) + c^{2}g''(x+ct) = c^{2} \frac{\partial^{2}}{\partial x^{2}} z(x,t)$

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