f(x)= so what does f'6=?

0 ⤊

f(x)= so what does f'6=?

f(x)= [x^7(x-6)^9]/(x^2+3)^7

f'6=?

2007-08-22 07:37:35 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics

4 answers

First find the derivative.

Then plug in 6 for all the remaining xes.

2007-08-22 07:41:10 · answer #1 · answered by Brian L 7 · 0⤊ 0⤋

I Dont Know But I Do Know That Alt+f4= Closing A Program

2007-08-22 07:45:18 · answer #2 · answered by jose g 1 · 0⤊ 0⤋

Zero

2007-08-22 07:47:05 · answer #3 · answered by Anonymous · 0⤊ 0⤋

f'(x) = (x^2+3)^7[(x^7*9(x-6)^8+ 7x^6(x-6)^9] - (x^7(x-6)^9(7x^2+2)^6(2x)/(x^2+3)^14
f'(6) = {(36+3)^7[6^7*9(6-6)^8+ 7*6^6(6-6)^9-6^7(6-6)^9(7*6^2+2)(2*6)]/(x^2+3)^14
f'(6) = 0

2007-08-22 08:02:42 · answer #4 · answered by ironduke8159 7 · 0⤊ 0⤋