Hello MathHelp.

So i'm trying to solve this problem below

Attachment 23105

And using the rules of differentiation that i know of i fail. Perhaps the book that i am reading is wrong. If anyone can help it would be greatly appreciated.

- Simon.

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- Dec 18th 2011, 03:59 PMSimonFrisendalHelp with differentiation incl. ln.
Hello MathHelp.

So i'm trying to solve this problem below

Attachment 23105

And using the rules of differentiation that i know of i fail. Perhaps the book that i am reading is wrong. If anyone can help it would be greatly appreciated.

- Simon. - Dec 18th 2011, 06:37 PMvincisonfireRe: Help with differentiation incl. ln.
Hi,

These is nothing special there. Just use the product rule together with the chain rule.

Note that you have something like $\displaystyle (x-1)ln(x-1)-x ln(x)$. If you differentiate you will have something of the form $\displaystyle ln(x-1)-{x-1\over x-1}-ln(x)+{x\over x} = ln(x-1)-ln(x)$.

It is true that all the constants makes it messy, but keep you head straight and just use the basics rules you know. - Dec 18th 2011, 06:46 PMpickslidesRe: Help with differentiation incl. ln.
First take out the constant

$\displaystyle \displaystyle \frac{d}{dE}\kappa_B\left( \left(1+\frac{E}{\epsilon}\right)\times \ln\left(1+ \frac{E}{\epsilon}\right)-\frac{E}{\epsilon}\ln\frac{E}{\epsilon}\right)$

$\displaystyle \displaystyle = \kappa_B\frac{d}{dE}\left( \left(1+\frac{E}{\epsilon}\right)\times \ln\left(1+ \frac{E}{\epsilon}\right)-\frac{E}{\epsilon}\ln\frac{E}{\epsilon}\right)$

then try the product rule term by term i.e.

$\displaystyle \displaystyle \frac{d}{dE}\left( \left(1+\frac{E}{\epsilon}\right)\times \ln\left(1+ \frac{E}{\epsilon}\right)\right) $

$\displaystyle \displaystyle = \frac{d}{dE}\left(1+\frac{E}{\epsilon}\right) \times \ln\left(1+ \frac{E}{\epsilon}\right)+\left(1+\frac{E}{\epsilo n}\right)\times \frac{d}{dE}\ln\left(1+ \frac{E}{\epsilon}\right)$ - Dec 19th 2011, 12:05 AMSimonFrisendalRe: Help with differentiation incl. ln.
Thank you pickslides :) That's easier than expected.

- Simon.