differentiation

As follows:


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2010-09-10 00:49:24 補充：
Just factor out (x+1) from (x-1)[2(x+1)]+(x+1)^2

2010-09-10 11:05:38 補充：
You write it wrongly.
Not (x-1)[2(x+1)+(x+1)^2], should be (x-1)[2(x+1)]+(x+1)^2.

2010-09-10 13:10:37 補充：
Correction from the last three line:
h'(x)=3x²+2x-1-(x+1)[2(x-1)+(x+1)]
h'(x)=3x²+2x-1-(x+1)(3x-1)
h'(x)=3x²+2x-1-(3x²+2x-1)
h'(x)=0

Or you can do it in a simpler way: Just expand h(x) first.

how to rearrange from { (x-1)[2(x+1)+(x+1)^2} to 2(x+1)[(x-1)+(x+1)]???

2010-09-10 09:09:15 補充：
When i do the factorization of (x-1)[2(x+1)+(x+1)^2]. i get (x-1)(x+1)[2+(x+1)]
anything wrong with it????

2010-09-10 12:09:01 補充：
(x-1)[2(x+1)]+(x+1)^2
factor out (x+1) : x+1[(x-1) [2]+(x+1)]
how can i reach to 2(x+1)[(x-1)+(x+1)]?? Put [2] in the front of x+1??? Is it possible??
thx