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

The derivative of f(x)=ln((e^x)/(1+e^x)) is 1-(e^x)/(1+e^x)

What is the first derivative of f(x)=ln((e^x)/(1+e^x)) ?

The first derivative of f(x)=ln((e^x)/(1+e^x)) is 1-(e^x)/(1+e^x)

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derivative of

f (x) = ln (\frac{e ^{x}}{1 + e ^{x}})

1 - \frac{e ^{x}}{1 + e ^{x}}

Solution steps

\frac{d}{d x} (ln (\frac{e ^{x}}{1 + e ^{x}}))

Simplify

ln (\frac{e ^{x}}{1 + e ^{x}}) : x - ln (1 + e^{x})

= \frac{d}{d x} (x - ln (1 + e^{x}))

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d x}{d x} - \frac{d}{d x} (ln (1 + e^{x}))

\frac{d x}{d x} = 1

\frac{d}{d x} (ln (1 + e^{x})) = \frac{e ^{x}}{1 + e ^{x}}

= 1 - \frac{e ^{x}}{1 + e ^{x}}

What is the derivative of f(x)=ln((e^x)/(1+e^x)) ?
The derivative of f(x)=ln((e^x)/(1+e^x)) is 1-(e^x)/(1+e^x)
What is the first derivative of f(x)=ln((e^x)/(1+e^x)) ?
The first derivative of f(x)=ln((e^x)/(1+e^x)) is 1-(e^x)/(1+e^x)