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Mathematics 26 Online
OpenStudy (anonymous):

calculus help! can some check my answer to the following q? find the derivative of the function.......y=(1+e^x/1-e^x)

OpenStudy (anonymous):

i mean y=ln(1+e^x/1-e^x)

OpenStudy (anonymous):

i got y(prime)=1/x

OpenStudy (anonymous):

Seperate the logarithm using the log properties so y= ln(1+e^x) - ln(1-e^x) the derivative of ln(x) = 1/(u) *u'

OpenStudy (anonymous):

derivative of ln(u) = 1/(u) * u'

OpenStudy (anonymous):

u lost me

OpenStudy (anonymous):

dido got it

OpenStudy (anonymous):

Okay I will solve it...

OpenStudy (anonymous):

well i dont and i am the one w. the question

OpenStudy (anonymous):

\[\frac{d}{dx}\ln(f(x))=\frac{f'(x)}{f(x)}\]

OpenStudy (anonymous):

NO! i NEED to NOW this STUFF!

OpenStudy (anonymous):

know it now

OpenStudy (anonymous):

i mean KNOW!

OpenStudy (anonymous):

i now you meant that you got it? take the derivative of the inside piece, divide it by the inside piece. that is all

OpenStudy (anonymous):

y' = (1/(1+e^x)) *e^x +(1/(1-e^x))* (-e^x)) y' = (e^x)/(1+e^x) + (-e^x)/(1-e^x)

OpenStudy (anonymous):

you can just about do it in your head with a little practice

OpenStudy (anonymous):

I won't bother simplifying that.

OpenStudy (anonymous):

thanks 4 making fun of me satellite 73 :p

OpenStudy (anonymous):

you know i am only funnin

OpenStudy (anonymous):

Funning haha...

OpenStudy (anonymous):

you got this or no?

OpenStudy (anonymous):

lololohaha but no seriously i dont understand it:{

OpenStudy (anonymous):

ok lets do this one \[\ln(\sin(x))\] in your head you can write \[\frac{\cos(x)}{\sin(x)}\]

OpenStudy (anonymous):

the derivative of sine is cosine, that goes up top, sine goes in the bottom

OpenStudy (anonymous):

\[\ln(x^2+2x)\] derivative is \[\frac{2x+2}{x^2+2x}\] just like that

OpenStudy (anonymous):

derivative of inside piece up top, inside piece itself down below

OpenStudy (anonymous):

okay but no......wait! ur confusing the beJesus outta me :IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII

OpenStudy (anonymous):

ok i will shut up and wait

OpenStudy (anonymous):

no dont shut up........shut up while taking it too fats maybe going slower? explaining each step?

OpenStudy (anonymous):

fast i mean UGH!

OpenStudy (anonymous):

you are asked for the derivative of the log of something right? i mean something other than just \(x\)

OpenStudy (anonymous):

ok?

OpenStudy (anonymous):

so i will say you are asked for the derivative of \(\ln(f(x))\) i.e. the log of some other function of \(x\)

OpenStudy (anonymous):

so far so good?

OpenStudy (anonymous):

ehhhhhhhhhhhh i guess :I

OpenStudy (anonymous):

like for example \(\ln(1+e^x)\) or \(\ln(\sin(x)\) or \(\ln(x^2+2x)\)

OpenStudy (anonymous):

in other words the log of something \(\ln(\text{something})\)

OpenStudy (anonymous):

we good so far?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

check your work on: wolframalpha.com

OpenStudy (anonymous):

ok so this is how you do it: make a fraction take the derivative of the inside piece, put it up top put the inside piece itself in the denominator

OpenStudy (anonymous):

okay

OpenStudy (anonymous):

that is all

OpenStudy (anonymous):

so how about \[\ln(\sin(x))\]?

OpenStudy (anonymous):

its ln(cosx/sinx)

OpenStudy (anonymous):

it would be without the log

OpenStudy (anonymous):

o.

OpenStudy (anonymous):

y?

OpenStudy (anonymous):

i can explain if you like, lets get a correct answer first

OpenStudy (anonymous):

k

OpenStudy (anonymous):

so?

OpenStudy (anonymous):

um. do i do the sma e4 my prob right??

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

same

OpenStudy (anonymous):

for my problem?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

after you simplify as suggested that is after you write \[\ln(1+e^x)-\ln(1-e^x)\]

OpenStudy (anonymous):

will i have e^x/1+e^x?

OpenStudy (anonymous):

for the first term, yes

OpenStudy (anonymous):

okay and i am supposed 2 do this as well to the second term?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

satellite 73?

OpenStudy (anonymous):

o okay and then do i do the quotient rule for both or wat?

OpenStudy (anonymous):

no then you are done

OpenStudy (anonymous):

what?! really?!

OpenStudy (anonymous):

you don't take the derivative twice! not unless you want the second derivative

OpenStudy (anonymous):

i think you are confused as to what we are doing we are not simplifying the expression, we are finding the derivative

OpenStudy (anonymous):

o okay thanks i think that is all for today thanks for helping me i really appreciate it...i dont have my brother to help me since he went back to school and its been pretty tough having that i am sucky at math THANK YOU!

OpenStudy (anonymous):

yw, hope it is clear, and also hope it is clear that it is easy to do! derivative of \(\ln(\sin(x))\) is \(\frac{\cos(x)}{\sin(x)}\) derivative of \(\ln(x^2+2x)\) is \(\frac{2x+2}{x^2+2x}\) derivative of \(\ln(1+e^x)\) is \(\frac{e^x}{1+e^x}\)

OpenStudy (anonymous):

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OpenStudy (anonymous):

lol

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