Given a shape function $f\left( {x,y} \right)$, where $x\left( {\xi ,\eta } \right)$ and $y\left( {\xi ,\eta } \right)$ are physical coordinates, and $\xi$ and $\eta$ are natural coordinates.
In order to find 1st partial derivative $\frac{{\partial f\left( {x,y} \right)}}{{\partial \xi }}$, obviously I have to apply chain rule
$$\frac{{\partial f\left( {x,y} \right)}}{{\partial \xi }} = \frac{{\partial x}}{{\partial \xi }}\frac{{\partial f}}{{\partial x}} + \frac{{\partial y}}{{\partial \xi }}\frac{{\partial f}}{{\partial y}}$$
Similarly, the 2nd partial derivative is equal
\begin{align} \frac{{{\partial ^2}f\left( {x,y} \right)}}{{\partial {\xi ^2}}} & = \frac{\partial }{{\partial \xi }}\left(\frac{{\partial f}}{{\partial \xi }}\right) = \frac{\partial }{{\partial \xi }}\left( {\frac{{\partial x}}{{\partial \xi }}\frac{{\partial f}}{{\partial x}} + \frac{{\partial y}}{{\partial \xi }}\frac{{\partial f}}{{\partial y}}} \right) \\ & = \frac{{{\partial ^2}x}}{{\partial {\xi ^2}}}\frac{{\partial f}}{{\partial x}} + \frac{{\partial x}}{{\partial \xi }}\left( {\frac{{\partial x}}{{\partial \xi }}\frac{{{\partial ^2}f}}{{\partial {x^2}}} + \frac{{\partial y}}{{\partial \xi }}\frac{{{\partial ^2}f}}{{\partial x\partial y}}} \right) \\ & + \frac{{{\partial ^2}y}}{{\partial {\xi ^2}}}\frac{{\partial f}}{{\partial y}} + \frac{{\partial y}}{{\partial \xi }}\left( {\frac{{\partial x}}{{\partial \xi }}\frac{{{\partial ^2}f}}{{\partial x\partial y}} + \frac{{\partial y}}{{\partial \xi }}\frac{{{\partial ^2}f}}{{\partial {y^2}}}} \right) \end{align}
But when I try to find 3rd derivative I get stuck on this stage
\begin{align} \frac{{{\partial ^3}f\left( {x,y} \right)}}{{\partial {\xi ^3}}} & = \frac{\partial }{{\partial \xi }}\left( {\frac{\partial }{{\partial \xi }}\left( {\frac{{\partial f}}{{\partial \xi }}} \right)} \right) = \frac{\partial }{{\partial \xi }}\left( {\frac{\partial }{{\partial \xi }}\left( {\frac{{\partial x}}{{\partial \xi }}\frac{{\partial f}}{{\partial x}} + \frac{{\partial y}}{{\partial \xi }}\frac{{\partial f}}{{\partial y}}} \right)} \right) \\ & = \frac{\partial }{{\partial \xi }}\left( {\frac{{{\partial ^2}x}}{{\partial {\xi ^2}}}\frac{{\partial f}}{{\partial x}} + \frac{{\partial x}}{{\partial \xi }}\frac{{\partial x}}{{\partial \xi }}\frac{{{\partial ^2}f}}{{\partial {x^2}}} + \frac{{\partial x}}{{\partial \xi }}\frac{{\partial y}}{{\partial \xi }}\frac{{{\partial ^2}f}}{{\partial x\partial y}}} \right) \\ & + \frac{\partial }{{\partial \xi }}\left( {\frac{{{\partial ^2}y}}{{\partial {\xi ^2}}}\frac{{\partial f}}{{\partial y}} + \frac{{\partial y}}{{\partial \xi }}\frac{{\partial y}}{{\partial \xi }}\frac{{{\partial ^2}f}}{{\partial {y^2}}} + \frac{{\partial y}}{{\partial \xi }}\frac{{\partial x}}{{\partial \xi }}\frac{{{\partial ^2}f}}{{\partial x\partial y}}} \right) \end{align}
since calculation of $\frac{\partial }{{\partial \xi }}\left( {\frac{{{\partial ^2}f}}{{\partial {x^2}}}} \right)$, $\frac{\partial }{{\partial \xi }}\left( {\frac{{{\partial ^2}f}}{{\partial {y^2}}}} \right)$ and $\frac{\partial }{{\partial \xi }}\left( {\frac{{{\partial ^2}f}}{{\partial {x}}{\partial {y}}}} \right)$ looks tricky for me.
Could you help me please with calculation of 3rd derivative or, to be more exact, with calculation of $\frac{\partial }{{\partial \xi }}\left( {\frac{{{\partial ^2}f}}{{\partial {x^2}}}} \right)$?
You'll find that the operator expression for the chain rule is more manageable for these kind of calculations. This looks like $$\frac{\partial}{\partial \xi}=\frac{\partial x}{\partial \xi} \frac{\partial}{\partial x}+\frac{\partial y}{\partial \xi} \frac{\partial}{\partial y}$$ With this, for example, (subscripts noting partial derivatives) $$\frac{\partial f_{xx}}{\partial \xi}=\frac{\partial} {\partial \xi}f_{xx}=\frac{\partial x}{\partial \xi} \frac{\partial f_{xx}}{\partial x}+\frac{\partial y}{\partial \xi} \frac{\partial f_{xx}}{\partial y}$$ $$=\frac{\partial x}{\partial \xi} f_{xxx}+\frac{\partial y}{\partial \xi} f_{xxy}$$