Question

DERIVATIVE

Answer 1

\[\frac{ 1 }{ 2 }(\frac{ 1 }{ 2 }\ln \frac{ x+1 }{ x-1 }+\tan^{-1} x)\]

Answer 2

please help

Answer 3

what do you have so far?

Answer 4

well i split the natural log into ln (x+1)-ln(x-1)

Answer 5

then i got stuck i tried getting the derivative but idk it didnt make sense

Answer 6

1/2 [ 1/2(1/x+1 -1/x-1) + 1/(1+x^2)^1/2

Answer 7

]

Answer 8

if you dont get it i cran write the equation

Answer 9

\[\frac{ 1 }{ 2 }[\frac{ 1 }{ 2 } (\ln \frac{ 1 }{ x+1 } - \ln \frac{ 1 }{ x-1 }) + \frac{ 1 }{ 1+x^2 }]\]

Answer 10

my bad, without the Ln's

Answer 11

yeah that looks right, you probably could clean it up a little.

\[\frac{ 1 }{ 2 }\left( \frac{ 1 }{ 2 }\left( \frac{ 1 }{ x+1 }-\frac{ 1 }{ x-1 } \right) +\frac{ 1 }{ x^2+1 }\right)\]
\[\frac{ 1 }{ 4(x+1) }-\frac{ 1 }{ 4(x-1) }+\frac{ 1 }{ x^2+1 }\]
\[\frac{ x-1-(x+1)+4 }{ 4(x^2+1) }\]

Answer 12

wait, scratch that last step.

Answer 13

Should be\[\frac{ (x-1)(x^2+1)-(x+1)(x^2+1)+4(x+1)(x-1) }{ 4(x+1)(x-1)(x^2+1) }\]

Answer 14

on the last comment you did, wouldnt it be this? \[\frac{ 1 }{ 2x^2+2 }\]

Answer 15

since you also multiply the half times the derivative of arctanx

Answer 16

yeah I forgot to distribute the 1/2 there

Answer 17

oh okay one sec. im wrting it as we go to understand

Answer 18

my book says \[\frac{ 1 }{ 1-x^4 }\]

Answer 19

yeah i got that i fixed the error with the 1/2 that didnt get multiplied and got that

Answer 20

thanks!

Answer 21

ok I wrote it out, the typing was too much

Answer 22

yeah thats what i got too!

Answer 23

awesome!

Answer 24

this problems are driving me crazy. i hate this chapter of calculus

Answer 25

thanks

Answer 26

you're welcome.