# Taking second derivative of a derivative

1. Jul 5, 2010

### nkinar

Hello--

I'm in the process of implementing a PML for FDTD modeling.

I would like to take the derivative of the partial derivative shown below, but I am uncertain with respect to how I might proceed.

$$$\frac{\partial }{{\partial x}} \to \frac{1}{{1 + \frac{{i\sigma \left( x \right)}}{\omega }}}\frac{\partial }{{\partial x}}$$$

Essentially what I would like to do is take the derivative of a partial derivative, and also deal with the $$${i\sigma \left( x \right)}$$$ term, which is a function of position $$x$$.

This would result in the calculation of $$$\frac{{\partial ^2 }}{{\partial x^2 }}$$$

Last edited: Jul 5, 2010
2. Jul 5, 2010

### nkinar

Perhaps this would be the way to take the second derivative :

$$$\frac{{\partial ^2 }}{{\partial x^2 }} \to -\left( {1 + \frac{{i\sigma \left( x \right)}}{\omega }} \right)^{ - 2} \left( {\frac{{\partial \sigma \left( x \right)}}{{\partial x}}\frac{i}{\omega }} \right)\frac{\partial }{{\partial x}} + \frac{{\partial ^2 }}{{\partial x^2 }}\left( {1 + \frac{{i\sigma \left( x \right)}}{\omega }} \right)^{ - 1}$$$

Last edited: Jul 5, 2010
3. Jul 5, 2010

### HallsofIvy

Staff Emeritus
Use the product rule:
$$\frac{\partial }{\partial x}\left((1+ \frac{i\sigma}{\omega})^{-1}$$$$\frac{\partial y}{\partial x}\right)$$$$= \frac{\partial (1+ \frac{i\sigma}{\omega})^{-1}}{\partial y}{\partial x}$$$$+ (1+ \frac{i\sigma}{\omega})^{-1}\frac{\partial^2 y}{\partial x^2}$$

Last edited: Jul 5, 2010