# How do you find the derivative of #ln((e^x)/(1+e^x))#?

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To find the derivative of ln((e^x)/(1+e^x)), you can use the quotient rule combined with the chain rule. The quotient rule states that for functions u(x) and v(x), the derivative of u(x)/v(x) is (u'(x)v(x) - u(x)v'(x))/(v(x))^2. Applying this rule and the chain rule to ln((e^x)/(1+e^x)), we get:

[1/(e^x/(1+e^x)) * ((1+e^x)*(e^x)' - e^x*(1+e^x)')]/(e^x/(1+e^x))^2

Simplify and compute derivatives:

= [1/(e^x/(1+e^x)) * ((1+e^x)*e^x - e^x*(1+e^x))/(1+e^x)^2]

= [1/(e^x/(1+e^x)) * (e^x + e^(2x) - e^x - e^(2x))]/(1+e^x)^2

= [1/(e^x/(1+e^x)) * (e^(2x) - e^(2x))]/(1+e^x)^2

= 0/(1+e^x)^2

= 0

So, the derivative of ln((e^x)/(1+e^x)) is 0.

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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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