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I'm having troubles calculating $\frac{\partial^2 u}{\partial x^2}$. I know that $$\frac{\partial u}{\partial x} = \cos \varphi \frac{\partial u}{\partial r}-\frac{\sin \varphi}{r} \frac{\partial u}{\partial \varphi}$$ $$\frac{\partial r}{\partial x}=\cos\varphi \quad \text{and}\quad \frac{\partial \varphi}{\partial x} = -\frac{\sin\varphi}{r}$$

The solution should be $$\frac{\partial^2 u}{\partial x^2} = \cos^2 \varphi \frac{\partial^2 u}{\partial r^2}-\frac{2\sin \varphi \cos \varphi}{r} \frac{\partial^2 u}{\partial r\partial\varphi }+\frac{\sin^2 \varphi }{r^2} \frac{\partial^2 u}{\partial\varphi^2 }+\frac{\sin^2 \varphi}{r} \frac{\partial u}{\partial r }+\frac{2\sin \varphi \cos \varphi}{r^2} \frac{\partial u}{\partial\varphi }$$

I get how the first $3$ terms came to be but I don't know how to get the last two terms, namely $\frac{\sin^2 \varphi}{r} \frac{\partial u}{\partial r }+\frac{2\sin \varphi \cos \varphi}{r^2} \frac{\partial u}{\partial\varphi }$.

Can anyone explain?

Quotenbanane
  • 1,594

1 Answers1

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They are due to the "product rule". In addition to differentiating $\frac{\partial u}{\partial r}$ and $\frac{\partial u}{\partial \phi}$ with respect to r and $\phi$ you have to differentiate the coefficients, $cos(\phi)$ and $-\frac{sin(\phi)}{r}$ as well.

user247327
  • 18,710