My F'(x) that I got is this; (3(x^3+4x+6)^2(3x+4)
For some reason the comp is saying it's wrong???!!
Is it?
2007-11-16 16:35:44 · 5 answers · asked by RedSparkle 1 in Science & Mathematics ➔ Mathematics
shouldn't it be 3(x^3+4x+6)^2(3x^2+4)??
you missed the x^2. f' of x^3 = 3x^2
2007-11-16 16:40:16 · answer #1 · answered by norman 7 · 1⤊ 0⤋
F'(x)=3((x^3+4x+6)^2)(3x^2+4)
2007-11-17 00:48:18 · answer #2 · answered by billanders38 2 · 0⤊ 0⤋
Your answer is correct except that 3x should be 3x^2. Possibly, the computer simplified the answer more than you did.
For example, you can multiply the 3 by the 3x^2 + 4
(9x^2 + 12)(x^3 + 4x + 6)^2
Beyond that, you could square the trinomial (x^3 + 4x + 6) and even multiply the result of that by 9x^2 + 12.
2007-11-17 00:41:26 · answer #3 · answered by Scott K 2 · 0⤊ 0⤋
Yes, its wrong.
You forgot to square the last x
It should be:
F'(x) = 3*[ (x^3 + 4x + 6)^2 ]*(3x^2 + 4)
2007-11-17 00:43:41 · answer #4 · answered by BB 2 · 0⤊ 0⤋
ur comp is wrong, so there, fix it!!!
2007-11-17 00:53:03 · answer #5 · answered by ray y 2 · 0⤊ 0⤋