2007-09-09 10:26:47 · 4 answers · asked by tlcangel289 1 in Science & Mathematics ➔ Mathematics
f (x) = (3x - 1)^4
f `(x) = 4 (3x - 1) ³ (3)
f `(x) = 12 (3x - 1) ³
2007-09-13 07:58:51 · answer #1 · answered by Como 7 · 1⤊ 0⤋
d/dx u^n = n (u)^(n - 1) d/dx (u)
use the rule above and you'll get:
y' = 4 * (3x - 1)^(4 - 1) * d/dx (3x - 1)
y' = 4 * (3x - 1)^3 * (3)
y' = 12 (3x - 1)^3
2007-09-09 17:39:17 · answer #2 · answered by 7 · 0⤊ 0⤋
Let u = (3x-1), so
y = u^4
du/dx = 3, and:
dy/du = 4u^3, since
dy/dx = du/dx* dy/du, then
dy/dx = 3*4u^3 so,
dy/dx = 3*4(3x-1)^3, so
dy/dx = 12(3x-1)^3.
Hope this helps, Twiggy.
2007-09-09 17:46:36 · answer #3 · answered by Twiggy 7 · 0⤊ 0⤋
we have to use the chain rule
y'[(f(x)^k] = k[f(x)]^(k-1) * y'[(f(x)]
f(x) = 3x-1
k=4
4[(3x-1)^(4-1)]*(3)
simplified, 12[(3x-1)^3]
2007-09-09 17:37:36 · answer #4 · answered by imitative 2 · 0⤊ 0⤋