# Differentiation Problem -

1. Jun 18, 2009

### tim85ruhruniv

I was working on some of my own equations and today i ended up with this differentiation thinghy, I never expected this in my equation but it just turned up :( so if there's anybody out there who loves to solve math please give this a try :)

maybe its too simple :) ... i am just having doubts about the order of the operations and their influences...

me too will try solving it in the mean time...

$$\frac{d}{d\theta}\left\{ \nabla\left(\underset{k}{\sum}\left[n_{k}+\theta n_{j}^{\star}\right]z_{k}^{2}\right)^{\frac{1}{2}}\right\} =?$$

$$$n_{k},n_{j}^{\star}$$$ are independent of $$\theta$$
$$$n_{k},n_{j}^{\star}$$$ depend on x,y,z (grad) (Actually these variables are concentrations defined at each point in our domain)
$$$z_{k}^{2}$$$ is a constant (actually the valence of the kth chemical species)

$$\theta$$ is also independent of the geometry (x,y,z)... i call it the ALIEN variable :) because afterwards i have to kill it by setting it to zero, that is after differentiation...

Thanks a lot,

Tim

Last edited: Jun 18, 2009
2. Jun 18, 2009

### tim85ruhruniv

Is this Right ???

$$$=-\frac{1}{2}\left(\underset{k}{\sum}\left[n_{k}+\theta n_{j}^{\star}\right]z_{k}^{2}\right)^{-\frac{3}{2}}n_{j}^{\star}\nabla\underset{k}{\sum}\left[n_{k}+\theta n_{j}^{\star}\right]z_{k}^{2}$ $+\frac{1}{2}\underset{k}{\sum}\left[n_{k}+\theta n_{j}^{\star}\right]z_{k}^{2}\nabla n_{j}^{\star}$$$

can someone confirm :)

Thanks a lot ....