- #1
contempquant
- 13
- 0
Homework Statement
I'm trying to solve a double derivative of a function of a function problem, but can't seem to be able to differentiate once. I have the formula given by the chain rule,
find [tex]\frac{d}{dz}{E}(\bar{r})[/tex] when [tex]{E}(\bar{r})={E_0}(\bar{r})e^{ikz}[/tex] and [tex]\bar{r}[/tex] is a function of z.
Homework Equations
I have used [tex]\frac{df}{dz}=\frac{dg}{df}\frac{df}{dz}h(x)+\frac{dh}{df}\frac{df}{dz}g(x)[/tex]
where [tex]g(x)={E_0}(\bar{r})[/tex] and [tex]h(x)=e^{ikz}[/tex]
and [tex]f=E(\bar{r})=E_{0}(\bar{r})[/tex]
The Attempt at a Solution
I have then substituted in for these giving,
[tex]\frac{d{E_0}(\bar{r})exp^{ikz}}{dz}[/tex]=[tex]\frac{\partial{E_0}(\bar{r})}{\partial{E}(\bar{r})}\frac{\partial{E}(\bar{r})}{\partial{z}}*e^{ikz}+\frac{\partial{e^{ikz}}}{\partial{E}(\bar{r})}\frac{\partial{E}(\bar{r})}{\partial{z}}*E_{0}(\bar{r})[/tex]
The problem is i don't know how to differentiate any of the individual partial differentials because [tex]\bar{r}=r(z)[/tex], r is dependent on z. Any ideas? i really don't know how to start this one, and when this is done i have to differentiate it again. I'm sure if i can find out how to do this once i can do the second derivative but can't seem to do this one.
apparently the final solution is,
[tex](\frac{\partial^{2}E_{0}}{\partial{z}^{2}}+2ik\frac{\partial{E_0}}{\partial{z}}-k^{2}E_0)e^{ikz}[/tex]