What is the derivative of #ln((e^x)/(1+e^x))#?

Answer 1

#1/(1+e^x)#

#y'=1/(e^x/(1+e^x)) * (e^x(1+e^x)-e^xe^x)/(1+e^x)^2=1/e^x * (e^x+e^(2x)-e^(2x))/(1+e^x)=#
#=1/e^x * e^x/(1+e^x)=1/(1+e^x)#
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Answer 2

#1/(1+e^x)#

You can bypass using the quoitient rule for derivatives by using the quotient rule for logarithms

#ln(a/b) = lna - lnb#

In your case, you will have

#ln(e^x/(1+e^x)) = lne^x - ln(1+e^x)#
Now all you have to do is use the chain rule twice, once for #lnu#, with #u=e^x#, and once for #lnv#, with #v= 1+e^x#.
#d/dx[lne^x - ln(1+e^x)] = d/dxlne^x - d/dxln(1+e^x)#

You have

#d/dx(lnu) = d/(du)lnu * d/dx(u)#
#d/dx(lnu) = 1/u * d/dxe^x = 1/color(red)(cancel(color(black)(e^x))) * color(red)(cancel(color(black)(e^x))) = 1#

and

#d/dx(lnv) = d/(dv)lnv * d/dx(v)#
#d/dx(lnv) = 1/v * d/dx(1+e^x) = 1/(1+e^x) * e^x#

The target derivative will thus be

#d/dxln(e^x/(1+e^x)) = 1 - e^x/(1+e^x) = (1 + color(red)(cancel(color(black)(e^x))) - color(red)(cancel(color(black)(e^x))))/(1+e^x) = color(green)(1/(1+e^x)#
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Answer 3

The derivative of ln((e^x)/(1+e^x)) is:

1/(1+e^x)

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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