second order multivariable derivative

Question

$ u=u(x,t)$, $x=x(x')$, $t=t(t')$

I know that $$ \frac{\partial u}{\partial x'}=\frac{\partial u}{\partial x}\frac{dx}{dx'}+\frac{\partial u}{\partial t}\frac{\partial t}{\partial x'}=\frac{\partial u}{\partial x}\frac{dx}{dx'}$$

because $t$ is independent of $x'$

However, I don't know how to calculate $\dfrac{\partial^2u}{\partial x'^2}$

I've tried saying that

$$ \frac{\partial^2 u}{\partial x'^2} =\frac{\partial}{\partial x'}\bigg(\frac{dx}{dx'}\bigg)\frac{\partial u}{\partial x}+\frac{\partial}{\partial x'}\bigg(\frac{\partial u}{\partial x}\bigg)\frac{dx}{dx'}$$ $$=\frac{\partial}{\partial x'}\bigg(\frac{\partial u}{\partial x}\bigg)\frac{dx}{dx'}$$

as $\dfrac{\partial}{\partial x'}\bigg(\dfrac{dx}{dx'}\bigg)=0 $

but this leaves me with another expression I don't know how to solve

Answer 1

We have \begin{eqnarray*} \frac{\partial u}{\partial x'}=\frac{\partial u}{\partial x}\frac{\partial x}{\partial x'} \end{eqnarray*} Differentiating again gives \begin{eqnarray*} \frac{\partial^2 u}{\partial x'^2} &=& \frac{\partial }{\partial x'} \left(\frac{\partial u}{\partial x}\frac{\partial x}{\partial x'} \right) \\ &=& \frac{\partial x}{\partial x'} \frac{\partial }{\partial x} \left(\frac{\partial u}{\partial x} \right)\frac{\partial x}{\partial x'} + \frac{\partial }{\partial x'} \left(\frac{\partial x}{\partial x'} \right) \frac{\partial u}{\partial x} \\ &=& \left(\frac{\partial x}{\partial x'}\right)^2 \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 x}{\partial x'^2} \frac{\partial u}{\partial x}. \\ \end{eqnarray*}