# Derivative of log/exponential functions

1. Mar 27, 2012

### agenttiny200

1. The problem statement, all variables and given/known data

Find the derivative of the following function:
(2x)/(ex)

2. Relevant equations

f'(ax) = ax . lna
f(ex) = ex

I already know the answer, but I don't understand how it comes to it. why does the log(2/e) (from below, 'The attempt at a solution' step 3) appear? and why does it turn into {log(2)-1}? from my understanding, it should turn into log(2) - log(e-1)

3. The attempt at a solution

f(x) = (2x)/(ex)
f(x) = (2/e)x
f'(x) = {(2/e)x} . {log(2/e)}
f'(x) = {(2/e)x} . {log(2)-1}

2. Mar 27, 2012

### Dick

No, the answer is ok. log(a/b)=log(a)-log(b) so log(2/e)=log(2)-log(e).

3. Mar 27, 2012

### agenttiny200

But it is exactly with this that I am getting stuck, why does log(e) turn into 1? isn't it equal to 0.43429...?

4. Mar 27, 2012

### Dick

So you got it? log to the base 10 of e is 0.43429... But that's not what you want here, right?

5. Mar 27, 2012

### agenttiny200

well, to explain it better, I used this site:
http://www.wolframalpha.com/input/?i=derivative+calculator&f1=%282^x%29%2F%28e^x%29&x=0&y=0&f=Derivative.derivativefunction_%282^x%29%2F%28e^x%29&a=*FVarOpt.1-_**-.***Derivative.derivativevariable---.*--

If you click on 'show steps', the website explain how to solve it.

But I am not the sort to mindlessly copy down answers without understanding them, and I am very stuck on how log(2/e) turns into log(2)-1 when it should be log(2)-log(e). And I don't think this site is wrong, because I have used it for a week now for a lot of derivatives, and it has been on the dot so far.

P.S., sorry for the late reply, was occupied.

6. Mar 27, 2012

### emailanmol

I hope it's clear now.

ln is to the base e where as log is to the base 10.

Also log(a/b)= log(a)-log(b)

Not log(a)-log(1/b).

This rule can be derived using basic definition of log function.

NOTE :I HAVE EDITED THE QUOTE A LITTLE

Last edited: Mar 28, 2012
7. Mar 27, 2012

### agenttiny200

okay, maybe the site isn't all that good with log derivatives...I get how it becomes -1 now, thanks!

8. Mar 28, 2012

### scurty

The site is fine. Look at the little remark that it says in the first box.

"log(x) is the natural logarithm"

9. Mar 28, 2012

### emailanmol

Hey, don't think it that way.
I think Wolfram Alpha is the best site for mathematic calculations.It can do integrations as well

In fact in your answer, look at the first block at the lower right corner.Tou can see that its stated log is natural logarithm :-)

10. Mar 28, 2012

### emailanmol

Same pinch :P
I said the same thing :-)

11. Mar 28, 2012

### agenttiny200

actually, I just noticed that myself and was about to come back here to correct myself. I swear, this wolframalpha seems to be able to do anything mathematical.

12. Mar 28, 2012

### ShayanJ

Sorry if the thread is solved but I guess if I write exact calculations,The OP will be better.

$\frac{d}{dx} (\frac{2}{e})^x = \frac{d}{dx} e^{x \ln{\frac{2}{e}}}=\frac{d}{dx} e^ {x (\ln{2}-\ln{e})}=\frac{d}{dx} e^ {x (\ln{2}-1)}=e^{x(\ln{2}-1)} (\ln{2}-1)=(\ln{2}-1) (\frac{2}{e})^x$