let k(x) = 4x(5 + 2x)^9
2007-03-06 20:52:42 · 5 answers · asked by bballchic601 1 in Science & Mathematics ➔ Mathematics
derivative of k(x)=20(2x+5)^8*(4x+1)
TO THE SMART GUY THAT GAVE ME A "BAD ANSWER"
Using the chain rule:
k'(x)=4(5+2x)^9 + 4x*9(5+2x)^8 *2
=4(5+2x)^9 + 72x(5+2x)^8
Putting (5+2x)^8 as common factor:
k'(x)=20(2x+5)^8*(4x+1)
So my answer is very very right.
Next time think before.
2007-03-06 21:04:40 · answer #1 · answered by diamond 3 · 0⤊ 1⤋
k' (x) = 9(4x(5+2x))^8 + 4(5+2x) + 2
= 9(4x(5+2x))^8 + 22+ 8x
2007-03-07 06:22:43 · answer #2 · answered by sophia 2 · 0⤊ 0⤋
k'(x)=4(5+2x)^9 + 4x*9(5+2x)^8 *2
=4(5+2x)^9 + 72x(5+2x)^8
2007-03-07 05:06:33 · answer #3 · answered by SS 2 · 0⤊ 0⤋
k(x)=4x(5+2x)^9
=4x(2*9(5+2x)^8)+4(5+2x)^9
=4(5+2x)^8 *(1+18(5+2x))
=4(5+2x)^8*(91+36x)
=(364+144x) (5+2x)^8
2007-03-07 05:07:07 · answer #4 · answered by ammori 2 · 0⤊ 0⤋
use chain rule
2007-03-07 05:03:02 · answer #5 · answered by John 5 · 0⤊ 0⤋