Find the derivative of: 1/1 + e^-x

  • Thread starter roam
  • Start date
  • #1
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Hi,
Here's my question:
Find the derivative of:
1/1 + e^-x




Here is my attemp, since 1/x=lnx and -x=-1, => ln(1-e^-x)
I don't know if this is right, I would appreciate some guidance.
 

Answers and Replies

  • #2
1,631
4
well i am assuming you meant the following

[tex](\frac{1}{1+e^{-x}})'[/tex]

you need to apply the quotient rule here, do you know the quotient rule?

here it is in case you have forgotten it:

[tex](\frac{f}{g})'=\frac{f'g-g'f}{g^{2}}[/tex]
 
  • #3
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Or you can write the function as: [tex](1+e^{-x})^{-1}[/tex]

And then apply the power rule: [tex]\frac{d}{dx} u^n=nu^{n-1} \frac{du}{dx}[/tex]
 
  • #4
1,266
11
Ok, using the quotient rule:
f=1
f'=0
g=(1+e^-x)
g'=-e^-x

So we have:

http://img406.imageshack.us/img406/5082/sdfth1nz7.gif [Broken]

Is that right?


But using the other method I get:
(1+e^-x)^-1
So: (-e^-x) . -1 . (1+e^-x)^-2
inner function is: -e^-x

Which answer do you think is right?

Thank you.
 
Last edited by a moderator:
  • #5
1,631
4
\!\(\(0\[Cross]\((1 + e\^\(-x\))\) - \ \(-e\^\(-x\)\ \[Cross]\ 1\)\)\/\((1 + \
e\^\(-x\)\ )\)\^2\)
what on earth is this?
 
  • #6
CompuChip
Science Advisor
Homework Helper
4,302
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That's
[tex]\frac{0\times \left(e^{-x}+1\right)+e^{-x}\times
1}{\left(e^{-x}+1\right)^2}[/tex]

roam, next time please use TeXForm and put it between [tex], [/ tex] tags (without the space). And if you have Mathematica anyway, you can use it to check the answer:
FullSimplify[....youranswer... == D[ function, x ]]
 
  • #7
1,266
11
Thank you Compuchip, I understand it now and by simplifying it we get:
e^-x / (1+e^-x)^2

But I have another question:
http://img170.imageshack.us/img170/1766/xcvdxf1fs3.gif [Broken]
Let y = tan-1x
Then, Let x = tany
dx/dy = sec2y = 1+tan2y [Trigonometric identity]
Therefore dy/dx = 1/(1+tan2y) = 1/(1+x2)

Am I on the right track?


Thank you. :smile: :smile:
 
Last edited by a moderator:
  • #8
CompuChip
Science Advisor
Homework Helper
4,302
47
Not really. First of all, [itex]\tan^{-1}(\cdots)[/itex] can mean either [itex]1/\tan(\cdots)[/itex], so evaluate the tan function and then take its reciprocal. Or it can mean [itex]\arctan(\cdots)[/itex], which is the function such that if you take tan(arctan(x)) you get x back.

Since the derivative of the second one is hard do find by the methods you have been using so far, I'll just assume you meant to write
[tex]y = \frac{1}{\tan(\ln x)}[/tex]
For this, I'd suggest using the chain rule:
[tex]u(x) = \tan(\ln x) \qquad\Rightarrow\qquad y = u^{-1}, \frac{dy}{dx} = \left(\frac{dy}{du} \right) \left( \frac{du}{dx} \right).[/tex]
 
Last edited:
  • #9
Vid
401
0
y = arctan(ln(x))
tan(y) = ln(x)
Use implicit differentiation and solve for dy/dx.
 

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