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2007-11-22 07:28:11 · 6 answers · asked by a89411389 1 in Science & Mathematics ➔ Mathematics
let u = x^2...............then y = e^ u
du/dx = 2x................dy/du = e^u
chain rule: dy/dx = dy/du * du/dx
y = 2xe^(x^2)
2007-11-22 07:35:08 · answer #1 · answered by ? 3 · 2⤊ 1⤋
use chain rule
let u = x^2 du/dx = 2x
y = e^u dy/du = e^u
so dy/dx = 2xe^(x^2)
2007-11-22 07:35:02 · answer #2 · answered by norman 7 · 0⤊ 0⤋
y = e^(x^2)
dy/dx = e^(x^2)*d(x^2)
dy/dx = e^(x^2) (2x)
dy/dx = (2x) e^(x^2)
2007-11-22 07:35:41 · answer #3 · answered by mohanrao d 7 · 0⤊ 0⤋
dy/dx= 2x e^(x^2)
in general: y= e^f(x)
dy/dx= f'(x)e^f(x)
2007-11-22 07:44:46 · answer #4 · answered by Anonymous · 1⤊ 0⤋
using the chain rule, if u is a function of x, d(e^u)/dx = d(e^u)/du • du/dx, so
d(e^(x²))/dx = e^(x²) • 2x = 2xe^x²
2007-11-22 07:35:42 · answer #5 · answered by Philo 7 · 0⤊ 0⤋
dy/dx=(e^(x^2))(2x)
2007-11-22 07:34:58 · answer #6 · answered by ss89 2 · 0⤊ 0⤋