Exam Qs: Show that the derivative of ln(1-e^x) is 1/(Sqrt of 1 +2x) Watch

Tangeton
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So I used the chain rule... u = 1- e^-x, and so du/dx = e^-x
y = ln u, and so dy/du = 1/u.

Chain Rule: dy/dx = e^-x * 1/(1-e^-x) = e^-x/1-e^x

How does that make 1/e^x - 1? My thought was to say e^-x divided by -e^-x is 1/-1 and so that would make an answer of 1/1-1 = 1/0. How does it become 1/ e^x - 1?

Thank you.

*EDIT: It's not 1/Sqrt of 1+2x. Its 1/e^x - 1.
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Notnek
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(Original post by Tangeton)
So I used the chain rule... u = 1- e^-x, and so du/dx = e^-x
y = ln u, and so dy/du = 1/u.

Chain Rule: dy/dx = e^-x * 1/(1-e^-x) = e^-x/1-e^x

How does that make 1/e^x - 1? My thought was to say e^-x divided by -e^-x is 1/-1 and so that would make an answer of 1/1-1 = 1/0. How does it become 1/ e^x - 1?

Thank you.
Multiply top and bottom of the fraction by e^x.

I'm a bit confused why you mentioned 1/(Sqrt of 1 +2x) in your question.
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james22
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(Original post by Tangeton)
So I used the chain rule... u = 1- e^-x, and so du/dx = e^-x
y = ln u, and so dy/du = 1/u.

Chain Rule: dy/dx = e^-x * 1/(1-e^-x) = e^-x/1-e^x

How does that make 1/e^x - 1? My thought was to say e^-x divided by -e^-x is 1/-1 and so that would make an answer of 1/1-1 = 1/0. How does it become 1/ e^x - 1?

Thank you.
What do you think 1-e^-x divided by e^-x is?
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Tangeton
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(Original post by notnek)
Multiply top and bottom of the fraction by e^x.

I'm a bit confused why you mentioned 1/(Sqrt of 1 +2x) in your question.
Ohh my bad sorry I didn't double read my post, that's part of another question. It was meant to 1/(e^x-1).

Thank you for the quick response. But I am still a bit confused like... how do I know when to multiply, and why am I allowed to multiply? Because I thought you can't multiply something because you change its value if you do. So why in this case you're allowed? Is it because it is a fraction?
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Tangeton
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(Original post by james22)
What do you think 1-e^-x divided by e^-x is?
Its the other way around - e^-x divided by 1-e^-x and that's 1/1-1 = 1/0?
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james22
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(Original post by Tangeton)
Its the other way around - e^-x divided by 1-e^-x and that's 1/1-1 = 1/0?
No it isn't. I can't see why you think it is.

Are you just cancelling the two e^-x terms?
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Phichi
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(Original post by Tangeton)
Its the other way around - e^-x divided by 1-e^-x and that's 1/1-1 = 1/0?
Not quite.
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Tangeton
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(Original post by james22)
No it isn't. I can't see why you think it is.

Are you just cancelling the two e^-x terms?
Okay yeah my bad,.. so its e^-x/1 * -(e^-x/e^-x) = e^-x/1 x 1/1 = e^-x

So... where did I go wrong here? Now I cancelled out the e^-x correctly, or didn't I?
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james22
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(Original post by Tangeton)
Okay yeah my bad,.. so its e^-x/1 * -(e^-x/e^-x) = e^-x/1 x 1/1 = e^-x

So... where did I go wrong here? Now I cancelled out the e^-x correctly, or didn't I?
You need to learn how to manipulate fractions before you try questions like this. No offense but you are failing at fairly basic GCSE level stuff here.

You have just said that \frac{a}{1-a}=a. Do you really think this is right?
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Tangeton
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(Original post by james22)
You need to learn how to manipulate fractions before you try questions like this. No offense but you are failing at fairly basic GCSE level stuff here.

You have just said that \frac{a}{1-a}=a. Do you really think this is right?
Yeah sorry not that I am going to pass this year but you don't need to point it out because I already know. If you're going to just judge then you might just as well not help me. Like, I know its hard not to laugh at how bad I am at maths but really if you can't say something nice don't say it at all...

I originally was doing e^-x/1 - e^-x/e^-x and so that would be e^-x - 1 so that would be 1/1-a = a/1 - a/a = a - 1. But this still isn't 1/e^x-1 as if I turn it so 1/e^x - 1/1 it is 1-1/e^x-1 = 0/e^x-1?

Please help without criticizing me.
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james22
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(Original post by Tangeton)
Yeah sorry not that I am going to pass this year but you don't need to point it out because I already know. If you're going to just judge then you might just as well not help me. Like, I know its hard not to laugh at how bad I am at maths but really if you can't say something nice don't say it at all...
If you can't do basic GCSE maths you shouldn't be doing this level yet. You have to go back and revise the previous topics first. I'm not insulting you or mocking you, I'm pointing out that you need to have a good grasp of the basics before you can do questions that this level.

I originally was doing e^-x/1 - e^-x/e^-x and so that would be e^-x - 1 so that would be 1/1-a = a/1 - a/a = a - 1. But this still isn't 1/e^x-1 as if I turn it so 1/e^x - 1/1 it is 1-1/e^x-1 = 0/e^x-1?

Please help without criticizing me.
I don't think I am the right person to help with maths at this level. But you seen to have the idea that \frac{a}{b+c}=\frac{a}{b}+\frac{  a}{c} This is not true.

(Original post by TenOfThem)
x
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TenOfThem
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(Original post by Tangeton)
So I used the chain rule... u = 1- e^-x, and so du/dx = e^-x
y = ln u, and so dy/du = 1/u.

Chain Rule: dy/dx = e^-x * 1/(1-e^-x) = e^-x/1-e^x

How does that make 1/e^x - 1? My thought was to say e^-x divided by -e^-x is 1/-1 and so that would make an answer of 1/1-1 = 1/0. How does it become 1/ e^x - 1?

Thank you.

*EDIT: It's not 1/Sqrt of 1+2x. Its 1/e^x - 1.
This all seems a little complicated

First ... Are you differentiating ln(1-e^x) or ln(1-e^(-x))

If it is just x then you have (-e^x)/(1-e^x) and that does not give the answer required

If it is -x then you have \dfrac{e^{-x}}{1-e^{-x}} and this seems more likely


Assuming that it is the latter ... Fractions are unchanged if you multiply the numerator and denominator by the same thing ... In this case multiply them both by e^x
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