Question

y=(1-x) tanh^-1(x)
would the inside function be (1-x)?

Answer 1

The inside function would be x, as it's part of the arctan(x)

Answer 2

y=(1-x) tanh^-1(x) would the inside function be (1-x)?

Answer 3

mkk. im assuming this is using the chain rule.

looks sooo funky though.

Answer 4

Well you can break it up if you prefer... \[\large y=(1-x) arctanh (x) = \] \[\large arctanh (x) - x artanh (x)\]

Answer 5

Doesn't simplify it a whole lot though, still probably messy.

Answer 6

urgh. ended up with \[sech^2(x) arc(\tanh(x))\]

Answer 7

This is what wolframalpha gets... http://www.wolframalpha.com/input/?i=derivative+%281-x%29arctanh%28x%29

Answer 8

why does it bring x-1 under arctanhx

Answer 9

I think it's just putting them over a common denominator, but I'm not sure your solution is correct. Where did sech^2x come from? Derivative of arctanhx is 1/(1-x^2): http://ltcconline.net/greenl/courses/106/explogtrig/hyperbolic.htm

Answer 10

Oh i think i see where you got sechx. See the bottom of that link, you can simplify it.

Answer 11

yea i used the wrong rule. its the product rule for this one. although i dont get how they end up with a postive (x+1)
i end up with 1-tanh^-1(x)/1 + (1-x)/(1-x)^2

Answer 12

I think you can put them over a common denominator to probably get what they got. If you click step by step solution, you can see something more like your solution.

Answer 13

Notice you can use a difference of two squares on the (1-x^2)