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

1. Apr 8, 2008

### 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.

2. Apr 8, 2008

### 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:

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

3. Apr 8, 2008

### suspenc3

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

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

4. Apr 8, 2008

### roam

Ok, using the quotient rule:
f=1
f'=0
g=(1+e^-x)
g'=-e^-x

So we have:

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: Apr 8, 2008
5. Apr 8, 2008

### sutupidmath

what on earth is this?

6. Apr 8, 2008

### CompuChip

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

roam, next time please use TeXForm and put it between $$, [/ tex] tags (without the space). And if you have Mathematica anyway, you can use it to check the answer: FullSimplify[....youranswer... == D[ function, x ]] 7. Apr 9, 2008 ### roam Thank you Compuchip, I understand it now and by simplifying it we get: e^-x / (1+e^-x)^2 But I have another question: Let y = tan-1x Then, Let x = tany dx/dy = sec2y = 1+tan2y [Trigonometric identity] Therefore dy/dx = 1/(1+tan2y) = 1/(1+x2) Am I on the right track? Thank you. 8. Apr 10, 2008 ### CompuChip Not really. First of all, $\tan^{-1}(\cdots)$ can mean either $1/\tan(\cdots)$, so evaluate the tan function and then take its reciprocal. Or it can mean $\arctan(\cdots)$, which is the function such that if you take tan(arctan(x)) you get x back. Since the derivative of the second one is hard do find by the methods you have been using so far, I'll just assume you meant to write [tex]y = \frac{1}{\tan(\ln x)}$$
For this, I'd suggest using the chain rule:
$$u(x) = \tan(\ln x) \qquad\Rightarrow\qquad y = u^{-1}, \frac{dy}{dx} = \left(\frac{dy}{du} \right) \left( \frac{du}{dx} \right).$$

Last edited: Apr 10, 2008
9. Apr 10, 2008

### Vid

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