ln[1+exp(-x)]?

Question

help me differentiate this. Thanks

if possible, may i see the ways in order to get the answer. Thanx

Answer 1

let y = 1 + e^-x

now we can differentiate this on both sides to get

dy / dx = -e^-x

but we are given say, z = ln(y) and we need to find dz/dx

here dz/dy = 1/y

so dz/dx = dz/dy * dy/dx (the rulebook)
= 1/y * -e^-x
= 1/(1+e^-x) * -e^-x (putting the value of y)

so we have, = - exp(-x)/(1+exp(-x)).....a fairly easy one

Answer 2

d/dx)(ln(1+exp(-x))
= 1/(1+exp(-x)) d/dx(1+exp(-x)) using d/dx(ln y) = 1/y dy/dx
= -exp(-x)/(1+exp(-x) using d/dx(exp(y)) = exp(y) dy/dx

Answer 3

1/[1+exp(-x)] *-exp(-x)

Answer 4

-x