Derivative of arctan[(1 - x) / (1 + x)] Simplification

[communitycoll]

Homework Statement

Find the derivative of arctan[(1 - x) / (1 + x)].

Homework Equations

Everything in the "Show Steps" section:
http://www.wolframalpha.com/input/?i=derivative+arctan[(1+-+x)+/+(1+++x)]

My problem is that I don't know how Wolfram manages to simplify:

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

^ which is also what I've managed to get,

to get ,

- 1 / (1+x^2)

The Attempt at a Solution

Everything you see Wolfram does pretty much.

 

[Muphrid]

It multiplies through by the (1+x)^2 on the right and then expands the squares.

But really, it might be more instructive to use the chain rule and do this by hand?

 

[genericusrnme]

try multiplying everything out in the denominator and see what happens

 

[alan2]

Just multiply out your denominator. You get -2/{(1+x)^2+(1-x)^2}=-2/{2+2x^2}=...

 

[communitycoll]

Tell me what to do next or what I've done wrong here:

just showing the denominator:

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

= 1 + 2x + x^2 + (1 - 2x + x^2)(1 + 2x + x^2)

= 1 + 2x + x^2 + 1 + 2x + x^2 - 2x - 4x^2 - 2x^3 + x^2 + 2x^3 + x^4

= 2 + x^4 - 2x^2 + 2x

 

[genericusrnme]

Tell me what to do next or what I've done wrong here:

just showing the denominator:

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

= 1 + 2x + x^2 + (1 - 2x + x^2)(1 + 2x + x^2)

= 1 + 2x + x^2 + 1 + 2x + x^2 - 2x - 4x^2 - 2x^3 + x^2 + 2x^3 + x^4

= 2 + x^4 - 2x^2 + 2x

The denominator you gave was
[1 + ((1 - x) / (1 + x))^2](1 + x)^2

take the (1+x)^2 inside the brackets
[(1 + x)^2 + (1 + x)^2((1 - x) / (1 + x))^2]

Go from there, I'm not sure how you started off with
"= 1 + 2x + x^2 + [(1 - 2x + x^2)(1 + 2x + x^2)(1 + 2x + x^2) / (1 + 2x + x^2)]"

 

[communitycoll]

Ah, never mind, I was thinking I'm meant to multiply the fraction in the denominator as if they were being added/subtracted, trying to multiply the numerators by the denominators of both the fraction and (1 + x)^2. I get it now. Thanks, I appreciate your patience : D