Question

find the derivative
y=ln((1+e^x)/(1-e^x))
plz help!!!

Answer 1

\[\Large y=\ln\left(\frac{1+e^x}{1-e^x}\right)\]This thingy?

Answer 2

yes @zepdrix

Answer 3

Hmm so i guess applying a rule of logs before we differentiate might help:\[\Large \color{orangered}{\ln\left(\frac{a}{b}\right)\quad=\quad \ln(a)-\ln(b)}\]Understand how we can use this rule? :o

Answer 4

We don't HAVE TO apply the rule if you don't want to :)
It just allows us to avoid doing the quotient rule which can be a bit of a pain.

Answer 5

it would be ln (1+e^x) - ln(1-e^x)
right?

Answer 6

yes, good.

Answer 7

So ummm remember the rule for differentiating ln x? :)

Answer 8

no, not really :(

Answer 9

Hmm well here is the scoop :O\[\Large (\ln x)'\quad=\quad \frac{1}{x}\]Hopefully it looks familiar :D

Answer 10

yes it does!! lol sorry xp

Answer 11

When the contents of the log is `more than just x` we need to remember to apply the chain rule!
\[\Large (\ln \color{royalblue}{stuff})'\quad=\quad \frac{1}{\color{royalblue}{stuff}}(\color{royalblue}{stuff})'\]

Answer 12

ok so (1/1+e^x)(1e^x) yes??

Answer 13

derivative of (1+e^x) gave you (e^x) when doing your chain?
Ok looks good for the first log!

Answer 14

yes I get e^x :o

Answer 15

\[\Large y=\ln(1+e^x)-\ln(1-e^x)\]
\[\Large \color{royalblue}{y'}\quad=\quad\frac{1}{1+e^x}(e^x)-\color{royalblue}{\left[\ln(1-e^x)\right]'}\]Ok so we took the derivative of that first log, how bout the other one? :O

Answer 16

(-1/1-e^x)(-e^x)
right?

Answer 17

@zepdrix ^^

Answer 18

Ah yes, good! :)

Answer 19

You could probably simplify the expression by combining the fractions, but mehhh whatever.. depends whether or not your teacher wants you to do that.

Answer 20

yey!!!!!!!!!! \(*_*)/

Answer 21

and that is it!!!!???:O lol

Answer 22

lol ya i guess so :)

Answer 23

u guess?? lol :)

Answer 24

Well there is simplification that `can` be done, but that's just the annoying stuff at the end :b

As far as the differentiation is concerned, yes we're done! :D
Just depends whether or not you want to simplify it.

Answer 25

I simplified it:) lol and thank u so much!!!! I appreciate it:)

Answer 26

yay \c:/

Answer 27

lol u should be my actual tutor, u know everythin!