# Differentiation with chain rule and e^x and goodness Watch

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Part of a question requires me to differentiate:

(e

which of course is

e

So, unless I am being completely stupid, is

(e

The first e

-e

no? Somehow, it is instead

(e

which really confuses me.

(e

^{x})/(1+e^{x})which of course is

e

^{x}(1+e^{x})^{-1}So, unless I am being completely stupid, is

(e

^{x}* e^{x}* -1)(1+e^{x})^{-2}The first e

^{x}coming from the original "outside" term, the second coming from the differential of the 1-e^{x }and the -1 coming from the power. Making the final differential-e

^{2x}(1+e^{x})^{-2}no? Somehow, it is instead

(e

^{x}) / (1+e^{x})^{2}which really confuses me.

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#2

(Original post by

(e

The first e

**BobFredric**)(e

^{x}* e^{x}* -1)(1+e^{x})^{-2}The first e

^{x}coming from the original "outside" term, the second coming from the differential of the 1-e^{x }and the -1 coming from the power.u = 1 + e^x

So:

y = e^x / (1 + e^x) turns into

y = ( u - 1 ) / u

After that you should be able to simplify and use dy/dx = dy/du * du/dx to get the answer.

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#3

(Original post by

Part of a question requires me to differentiate:

(e

which of course is

e

So, unless I am being completely stupid, is

(e

The first e

-e

no? Somehow, it is instead

(e

which really confuses me.

**BobFredric**)Part of a question requires me to differentiate:

(e

^{x})/(1+e^{x})which of course is

e

^{x}(1+e^{x})^{-1}So, unless I am being completely stupid, is

(e

^{x}* e^{x}* -1)(1+e^{x})^{-2}The first e

^{x}coming from the original "outside" term, the second coming from the differential of the 1-e^{x }and the -1 coming from the power. Making the final differential-e

^{2x}(1+e^{x})^{-2}no? Somehow, it is instead

(e

^{x}) / (1+e^{x})^{2}which really confuses me.

we have (e^x)/(1+e^x)

let u=e^x and v=(1+e^x)

so du/dx=e^x and dv/dx=e^x

d/dx = (v du/dx - u dv/dx) / v^2

=(e^x + e^2x - e^2x) / (1+e^x)^2

=(e^x) / (1+e^x)

hope this makes sense...

if you haven't learnt the quotient rule, there are other ways to do it, but it's worth learning as i find it's quite easy to make mistakes with the chain rule

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(Original post by

Not entirely sure how you've got that answer and I don't see where you've used the chain rule, but easiest way I can see of using the chain rule for this is to say:u = 1 + e^xSo:y = e^x / (1 + e^x) turns intoy = ( u - 1 ) / uAfter that you should be able to simplify and use dy/dx = dy/du * du/dx to get the answer.

**__Adam__**)Not entirely sure how you've got that answer and I don't see where you've used the chain rule, but easiest way I can see of using the chain rule for this is to say:u = 1 + e^xSo:y = e^x / (1 + e^x) turns intoy = ( u - 1 ) / uAfter that you should be able to simplify and use dy/dx = dy/du * du/dx to get the answer.

(Original post by

Apply the quotient rule here.

we have (e^x)/(1+e^x)

let u=e^x and v=(1+e^x)

so du/dx=e^x and dv/dx=e^x

d/dx = (v du/dx - u dv/dx) / v^2

=(e^x + e^2x - e^2x) / (1+e^x)^2

=(e^x) / (1+e^x)

hope this makes sense...

if you haven't learnt the quotient rule, there are other ways to do it, but it's worth learning as i find it's quite easy to make mistakes with the chain rule

**halfhearted**)Apply the quotient rule here.

we have (e^x)/(1+e^x)

let u=e^x and v=(1+e^x)

so du/dx=e^x and dv/dx=e^x

d/dx = (v du/dx - u dv/dx) / v^2

=(e^x + e^2x - e^2x) / (1+e^x)^2

=(e^x) / (1+e^x)

hope this makes sense...

if you haven't learnt the quotient rule, there are other ways to do it, but it's worth learning as i find it's quite easy to make mistakes with the chain rule

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#5

(Original post by

I don't remember ever being taught that version of the chain rule. I was always taught when y= a(b)^c, dy/dx = a*(dy/dx of b)*c(b)^(c-1).I guess the problem here, after thinking through, is that you

.

**BobFredric**)I don't remember ever being taught that version of the chain rule. I was always taught when y= a(b)^c, dy/dx = a*(dy/dx of b)*c(b)^(c-1).I guess the problem here, after thinking through, is that you

**can't use the chain rule I learnt with the outside term being in terms of x**. I guess I should do the product or quotient rule..

The chain rule is for when you have a function of another function, so if you had:

y = (1 + e^x )^-1 you could say

b = 1 + e^x. So you can write y as a function of b, which is a function of x.

y = b^-1

After that:

dy/dx = dy/db * db/dx <--- this bit is the definition of the chain rule

Where you wrote:

I was always taught when y= (b)^c, dy/dx = (dy/dx of b)*c(b)^(c-1)

dy/dx of b is called db/dx

Also b^c has changed into cb^(c-1), which is called dy/db

So dy/dx is dy/db * db/dx <-- The definition of the chain rule I gave above.

After a while you do start thinking "differentiate the bit in the bracket, then multiply by the power and reduce the bracket's power", but what you're actually doing there is making a substitution and applying the above rule.

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