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I have a partial derivative: $$\frac{\partial u}{\partial x}= \frac{\partial u}{\partial \zeta}+\frac{\partial u}{\partial \varepsilon}$$ I want to find the second order partial derivative $\frac{\partial ^2u}{\partial x^2}$, and online I found the answer as: $$\frac{\partial ^2u}{\partial x^2}=\frac{\partial ^2u}{\partial \varepsilon^2}+2\frac{\partial ^2u}{\partial \varepsilon\zeta}+\frac{\partial ^2u}{\partial \zeta^2}$$ Can someone explain how they arrived at this?

If it's of any relevance, this is dealing with the 1-D wave equation and it's coming from here (equations 6 and 7).

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Observe \begin{align} \frac{\partial}{\partial x} u = \left(\frac{\partial}{\partial \zeta}+\frac{\partial}{\partial \varepsilon} \right) u \end{align} then \begin{align} \frac{\partial^2}{\partial x^2} u =&\ \left(\frac{\partial}{\partial \zeta}+\frac{\partial}{\partial \varepsilon} \right) \left(\frac{\partial}{\partial \zeta}+\frac{\partial}{\partial \varepsilon} \right) u = \left(\frac{\partial}{\partial \zeta}+\frac{\partial}{\partial \varepsilon} \right)\left(\frac{\partial u}{\partial \zeta}+\frac{\partial u}{\partial \varepsilon}\right) \\ =&\ \frac{\partial}{\partial \zeta}\left(\frac{\partial u}{\partial \zeta}+\frac{\partial u}{\partial \varepsilon}\right)+\frac{\partial}{\partial \varepsilon}\left(\frac{\partial u}{\partial \zeta}+\frac{\partial u}{\partial \varepsilon}\right) \\ =&\ \frac{\partial^2u}{\partial\zeta^2}+2\frac{\partial^2 u}{\partial \zeta\partial \varepsilon} + \frac{\partial^2 u}{\partial \varepsilon^2} \end{align}

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  • $\begingroup$ Thanks for the help! Just a question, what exactly does it mean to write $\frac {\partial} {\partial \zeta}$ on its own (i.e. $\frac {\partial} {\partial \zeta}$ + $\frac {\partial} {\partial \varepsilon}$)? I understand $\frac {\partial u} {\partial \zeta}$. $\endgroup$ – Andi Gu Apr 1 '17 at 4:37
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    $\begingroup$ When I rewrite $\partial/\partial \zeta$ I am viewing the object as an operator which acts on functions $u$ (like a matrix acting on a vector, i.e. $A(x) = Ax$). $\endgroup$ – Jacky Chong Apr 1 '17 at 4:56

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