# Differentiation with chain rule and e^x and goodnessWatch

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#1
Part of a question requires me to differentiate:

(ex)/(1+ex)
which of course is
ex(1+ex)-1

So, unless I am being completely stupid, is

(ex * ex * -1)(1+ex)-2

The first ex coming from the original "outside" term, the second coming from the differential of the 1-ex and the -1 coming from the power. Making the final differential

-e2x(1+ex)-2

(ex) / (1+ex)2

which really confuses me.
0
4 years ago
#2
(Original post by BobFredric)
(ex * ex * -1)(1+ex)-2

The first ex coming from the original "outside" term, the second coming from the differential of the 1-ex and the -1 coming from the power.
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^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.
1
4 years ago
#3
(Original post by BobFredric)
Part of a question requires me to differentiate:

(ex)/(1+ex)
which of course is
ex(1+ex)-1

So, unless I am being completely stupid, is

(ex * ex * -1)(1+ex)-2

The first ex coming from the original "outside" term, the second coming from the differential of the 1-ex and the -1 coming from the power. Making the final differential

-e2x(1+ex)-2

(ex) / (1+ex)2

which really confuses me.
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
1
#4
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.
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.

(Original post by 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
Yes, I know the quotient rule - I just tend to go with chain and product rule unless it's super complicated because quotient rule tends to be more work for most cases. I understood the way the answer was met with the quotient rule but just wondered why my chain rule failed. But thanks for the response .
0
4 years ago
#5
(Original post by 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.
.
Oh, I see what you mean, if you had something like (1 + e^x)^-1 you could use the chain rule, the problem with your question is that it was 2 functions of x multiplied together, so you would have to use the product/quotient rule unless you made a substitution.

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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