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Want to find the second order of $$\frac{\partial z}{\partial x}=\frac{\partial z}{\partial g}\frac{\partial g}{\partial x}+\frac{\partial z}{\partial h}\frac{\partial h}{\partial x}$$

Normally if I have something like $$\frac{\partial z}{\partial x}=\frac{\partial z}{\partial u}+\frac{\partial z}{\partial v}$$ to find the second order I'd do $$\frac{\partial^2 z }{\partial x^2}=\bigg( \frac{\partial }{\partial u}+\frac{\partial }{\partial v} \bigg) \bigg( \frac{\partial z}{\partial u}+\frac{\partial z}{\partial v} \bigg)$$ But now you can't really do the same thing so is this what you do (use product rule) $$\frac{\partial ^2 z}{\partial x^2}=\frac{\partial z}{\partial g}\frac{\partial ^2 g}{\partial x^2} +\frac{\partial ^2 z}{\partial g^2}\frac{\partial g}{\partial x}\frac{\partial g}{\partial x}+\frac{\partial z}{\partial h}\frac{\partial ^2 h}{\partial x^2}+\frac{\partial ^2 z}{\partial h^2}\frac{\partial h}{\partial x}\frac{\partial h}{\partial x}$$

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  • $\begingroup$ If $z$ is a function of just $g$ and $h$ which are functions of $x$ then the last equation is correct. $\endgroup$ – Winther Apr 23 '15 at 18:44
  • $\begingroup$ It is $z=f(g(x,y),h(x,y))$. So yeah. $\endgroup$ – snowman Apr 23 '15 at 18:47

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