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

Answer 1

Ezekiel Alexander

# 1/(1+e^x)#

#y = ln((e^x)/(1+e^x)) = ln (u), \qquad u = (e^x)/(1+e^x)#

and we use the chain rule #dy/dx = dy/(du) * (du)/dx#

#dy/(du) = 1/u = (1+ e^x)/(e^x)#

from quotient rule:

#(du)/(dx) = (e^x (1+ e^x) - e^x * e^x)/(1+e^x)^2# # = (e^x )/(1+e^x)^2 #

#\implies (dy)/(dx) = (1+ e^x)/(e^x) * (e^x )/(1+e^x)^2 = 1/(1+e^x)#

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Answer 2

Charles Anctil

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

(d/dx) ln((e^x)/(1+e^x)) = (1 / ((e^x)/(1+e^x))) * (d/dx) ((e^x)/(1+e^x))

Using the quotient rule and the chain rule:

(d/dx) ((e^x)/(1+e^x)) = ((e^x)(1+e^x) - e^x(e^x)) / (1+e^x)^2

Simplify the expression:

= (e^x + e^x - e^(2x)) / (1+e^x)^2

= (2e^x - e^(2x)) / (1+e^x)^2

So, the derivative of ln((e^x)/(1+e^x)) with respect to x is:

= (2e^x - e^(2x)) / ((1+e^x)^2)

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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.