Derivative of log/exponential functions

[agenttiny200]

Homework Statement

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

Homework Equations

f'(a^x) = a^x . lna
f(e^x) = e^x

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)

The Attempt at a Solution

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

 

[Dick]

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

 

[agenttiny200]

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

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...?

why does the log(2/e) (from below, 'The attempt at a solution' step 3) appear?

And forgot about this question of mine, revised my properties of log derivatives and saw the answer.

 

[Dick]

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...?



And forgot about this question of mine, revised my properties of log derivatives and saw the answer.

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

 

[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.

 

[emailanmol]

f'(a^x) = a^x .lna
f(e^x) = e^x



f(x) = (2^x)/(e^x)
f(x) = (2/e)^x
f'(x) = {(2/e)^x} . {log(2/e)}
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(THE STEP SHOULDN'T HAVE LOG.IT SHOULD BE LN FUNCTION.)

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

 

[agenttiny200]

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

 

[scurty]

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---.*--

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

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

"log(x) is the natural logarithm"

 

[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 :-)

 

[emailanmol]

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

"log(x) is the natural logarithm"

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

 

[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.

 

[ShayanJ]

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

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