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

[roam]

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.

 

[sutupidmath]

well i am assuming you meant the following

(\frac{1}{1+e^{-x}})'

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

here it is in case you have forgotten it:

(\frac{f}{g})'=\frac{f'g-g'f}{g^{2}}

 

[suspenc3]

Or you can write the function as: (1+e^{-x})^{-1}

And then apply the power rule: \frac{d}{dx} u^n=nu^{n-1} \frac{du}{dx}

 

[roam]

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

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.

 

[sutupidmath]

\!\(\(0\[Cross]\((1 + e\^\(-x\))\) - \ \(-e\^\(-x\)\ \[Cross]\ 1\)\)\/\((1 + \
e\^\(-x\)\ )\)\^2\)

what on Earth is this?

 

[CompuChip]

That's
\frac{0\times \left(e^{-x}+1\right)+e^{-x}\times<br /> 1}{\left(e^{-x}+1\right)^2}

roam, next time please use TeXForm and put it between , [/ tex] tags (without the space). And if you have Mathematica anyway, you can use it to check the answer: <br /> FullSimplify[...youranswer... == D[ function, x ]]

 

[roam]

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
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.

 

[CompuChip]

Not really. First of all, \tan^{-1}(\cdots) can mean either 1/\tan(\cdots), so evaluate the tan function and then take its reciprocal. Or it can mean \arctan(\cdots), 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
y = \frac{1}{\tan(\ln x)}
For this, I'd suggest using the chain rule:
u(x) = \tan(\ln x) \qquad\Rightarrow\qquad y = u^{-1}, \frac{dy}{dx} = \left(\frac{dy}{du} \right) \left( \frac{du}{dx} \right).

 

[Vid]

y = arctan(ln(x))
tan(y) = ln(x)
Use implicit differentiation and solve for dy/dx.