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.
Differentiate f(x)= (8-2x)^3(x^2+1)^5.
My ans is 2(8-2x)^2(x^2+1)^4(-3x^2-10x+37). Plz recheck my simplify ans.
-my simplify ans is not the same as wolfram but the long ans is same with wolfram-
3 answers
I still don't 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 didn't get the same ans.
= (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 didn't 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.
(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.