Posted by Ifi on Sunday, October 28, 2012 at 4:44am.
f' = 3(8-2x)^2 (-2) (x^2+1)^5 + 5(8-2x)^3 (x^2+1)^4 (2x)
= (8-2x)^2 (x^2+1)^4 (-6(x^2+1) + 5(8-2x)(2x))
= (8-2x)^2 (x^2+1)^4 (26x^2 - 80x + 6)
= -8(4-x)^2 (x^2+1)^4 (13x^2 - 40x + 3)
which is what Wolfram gets.
Watch for those pesky chain rule factors and minus signs.
I still dont get it how do u simplify from
= (8-2x)^2 (x^2+1)^4 (26x^2 - 80x + 6) to
= -8(4-x)^2 (x^2+1)^4 (13x^2 - 40x + 3) ... ?
What can i conclude from it is (4-x)^2 &(13x^2 - 40x + 3) is divided by 2 . Then (x^2+1)^4 is not simplified by any no. Lastly, why do u put -8? if u try to expand it back it didnt get the same ans.
Really. Forgotten your algebra I?
(8-2x)^2 = 4(4-x)^2
The intervening two lines should have had a "-" sign in front, from the (-2) factor. My typo. No need to pull out the 8; just felt like it. Prolly should leave it as 8-2x, so it looks more like the original syntax.
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