How do you solve to find the derivative of a square root of x^2-2x+3?

0 ⤊

2007-05-09 17:08:30 · 3 answers · asked by Troopa06 1 in Science & Mathematics ➔ Mathematics

You use the chain rule...

f'(x)=(1/(2*sqrt(x^2-2x+3))*(2x-2)
=(2x-2)/(2*sqrt(x^2-2x+3)
=(x-1)/(sqrt(x^2-2x+3)

2007-05-09 17:56:06 · answer #1 · answered by theholeinyourculture 2 · 0⤊ 0⤋

use the chain rule.

Let u=x^2 - 2x +3, and f(x)= sqrt(x).

Then d/dx = f'(x)d/du

= [(1/2)u^(-1/2)] * [2x -2]

= [(1/2)(x^2 - 2x +3)^(-1/2)][2x -2].

2007-05-09 17:57:31 · answer #2 · answered by chancebeaube 3 · 0⤊ 0⤋

y = â x^2-2x+3
= (x^2-2x+3)^Â½

use the chain rule

dy/dx = Â½(x^2-2x+3)^-Â½ . (2x - 3)

2007-05-09 19:22:02 · answer #3 · answered by fred 5 · 0⤊ 0⤋