2007-07-02 03:18:35 · 4 answers · asked by Demetria S 1 in Science & Mathematics ➔ Mathematics
Do you mean the slope of the function f(x) = (x - 1) / (x + 1)? The derivative of this function is f'(x) = [(x + 1)(1) - (x - 1)(1)] / (x + 1)^2 = (x + 1 - x + 1) / (x^2 + 2x + 1) = 2 / (x^2 + 2x + 1), which I got using the quotient rule for differentiation (lo-d-hi minus hi-d-lo over lo-squared). When you evaluate this for x = 0, you should get a result of 2. So the slope is 2.
2007-07-02 03:24:56 · answer #1 · answered by DavidK93 7 · 1⤊ 0⤋
x = 1
2007-07-02 10:24:40 · answer #2 · answered by ah tah 1 · 0⤊ 2⤋
y(x) = (x-1) / (x+1)
use quotient rule to find derivative y' = slope
y' (x) = [ (1)(x+1) - (x-1)(1) ] / [ (x+1)^2 ]
y' (x) = 2 / (x+1)^2
substitute x = 0 to y' function, we got
slope = y' (0) = 2
2007-07-02 10:34:20 · answer #3 · answered by Feb 3 · 0⤊ 0⤋
y = (x-1)(x+1)^(-1) and using chain rule
y' = (x+1)^(-1) + (x-1)(-1)(x+1)^(-2)
simplifying
y' = (1/(x+1))[ (-(x-1)/(x+1)^2) + 1] = (2- x) / (x+1)^2
So evaluating y' at x = 0
y'(0) = 2/1 = 2 and this is the desired slope
2007-07-02 10:27:42 · answer #4 · answered by Bazz 4 · 0⤊ 1⤋