2/ [(e^-x)+1] dx=
2007-05-29 01:57:56 · 4 answers · asked by chetzel 3 in Science & Mathematics ➔ Mathematics
g(x) = e^-x+1
g'(x) = -e^-x
f(x) = 2/x
f'(x) = -2/x^2
(d/dx) f(g(x)) = f'(g(x))g'(x))
=(-2/((e^-x)+1)^2)(-(e^-x))
=2(e^-x)/((e^-x)+1)^2
Use the chain rule
2007-05-29 02:06:09 · answer #1 · answered by A confused bio student 2 · 0⤊ 1⤋
As the numerator is constant, you can rewrite this as:
2( e^(-x) + 1 )^(-1)
Now differentiate with respect to e^(-x) + 1, and multiply the result by the differential of e^(-x) + 1. This gives:
2(- ( e^(-x) + 1 )^(-2) ) * (-e^(-x) )
= 2e^(-x) / (e^(-x) + 1)^2.
2007-05-29 02:15:05 · answer #2 · answered by Anonymous · 0⤊ 1⤋
f(x) = 2/[e^(-x) +1] = 2 [e^(-x) + 1]^(-1). . So,
f'(x) = 2 (-1) [e^(-x) + 1]^(-2) (e^(-x) +1)' = -2[e^(-x) + 1]^(-2) (-e^(-x)) = 2(e^(-x)) [e^(-x) + 1]^(-2) which is the same as
f'(x) = (2 e^(-x))/(e^(-x) +1)^2
2007-05-29 02:33:31 · answer #3 · answered by Steiner 7 · 0⤊ 0⤋
2 /[(e^-x) + 1] can be written as 2[(e^-x) + 1]^-1
So derivative is (2(-1)[(e^-x) + 1]^-2)(e^-x)(-1)
or 2(e^-x) / [(e^-x) + 1]^2
2007-05-29 02:30:11 · answer #4 · answered by dongskie mcmelenccx 3 · 0⤊ 0⤋