2007-06-29 12:37:56 · 4 answers · asked by Kelly 1 in Science & Mathematics ➔ Mathematics
Hi,
We will use d/dx[u/v] = u'v - uv')/v²
f(x)= ((x-2) / (x+1))² = (x-2)² / (x+1)²
f'(x) = [2(x - 2)(x + 1)² - (x - 2)²(2)(x + 1)]/(x + 1)^4
f'(x) = 2(x - 2)(x + 1)[x + 1 - (x - 2)]/(x + 1)^4
f'(x) = 2(x - 2)(x + 1)[x + 1 - x + 2]/(x + 1)^4
f'(x) = 2(x - 2)(x + 1)(3)/(x + 1)^4
f'(x) = 6(x - 2)/(x + 1)³
I hope that helps!! :-)
2007-06-29 12:50:37 · answer #1 · answered by Pi R Squared 7 · 0⤊ 0⤋
f(x) = (x - 2)² / (x + 1)²
f `(x) is given by:-
[(x + 1)².2.(x - 2) - (x - 2)².2.(x + 1)] / (x + 1)^4
(x + 1).(x - 2).[ 2.(x + 1) - 2.(x - 2) ] / (x + 1)^4
(x - 2).[ 6 ] / (x + 1)³
6.(x - 2) / (x + 1)³
2007-07-03 05:24:14 · answer #2 · answered by Como 7 · 0⤊ 0⤋
chain rule & quotient rule
f'(x) = 2[(x-2) / (x+1)] { [(x+1)*1 - (x-2)*1] / (x+1)^2 }
= 2[(x-2) / (x+1)] { [3] / (x+1)^2 }
=6(x-2) / (x+1)^3
2007-06-29 14:41:37 · answer #3 · answered by qwert 5 · 0⤊ 0⤋
Chain rule and quotient rule within it.
f'(x)=2[(x-2)/(x+1)]*[(x+1)(1)-(x-2)(1)]/(x+1)²
=(2x-4)/(x+1)*3/(x+1)²
=6(x-2)/(x+1)³
mouseover to see past the ellipses.
2007-06-29 12:46:49 · answer #4 · answered by jsoos 3 · 0⤊ 0⤋