2006-11-01 00:23:35 · 5 answers · asked by sean c 1 in Science & Mathematics ➔ Mathematics
using chain rule and product rule
y'= 3x^(2)*tan(4x)
+4x^(3)*sec^(2)(4x)
= x^2(3tan(4x)+4x*sec^(2)(4x)
i hope that this helps
2006-11-01 03:26:58 · answer #1 · answered by Anonymous · 0⤊ 0⤋
Product rule and chain rule:
y' = 3x^2 tan 4x + 4x^3 sec^2 (4x)
2006-11-01 00:26:41 · answer #2 · answered by Clueless 4 · 1⤊ 0⤋
y=x^3tan4x
y'=(x^3)' tan(4x) + x^3 tan'(4x)
y'=3x^2 tan4x + x^3 *1/(cos4x)^2 * 4
I'm not sure but I am pretty sure this is the result
2006-11-01 00:54:28 · answer #3 · answered by gljivarm 2 · 0⤊ 0⤋
dydx = (X^3)' tan4x x (X^3) (tan4x)'
= (3X^2) tan 4x x (X^3) 4sec^2 x
= x^2 (3tan 4x * 4x sec^2 x)
2006-11-01 02:00:02 · answer #4 · answered by luv_phy 3 · 0⤊ 0⤋
use the product rule, (i just told you it) and the chain rule
chain rule is df/dx = df/du * du/dx
let f(x) = tan(4x) become f(u) = tan(u) where u=4x
2006-11-01 00:26:35 · answer #5 · answered by Stuart T 3 · 1⤊ 0⤋