derivative f(x)=ln((e^x)/(1+e^x))

Lösung

ableitung von

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

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

Schritte zur Lösung

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

Vereinfache

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

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

Wende die Summen-/Differenzregel an:

(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}}

d er i v a t i v e y = ln (\frac{3}{x})

n = 0 \sum \infty \frac{2}{n + 1}

\int tan^{5} (4 x) sec^{3} (4 x) d x

n = 1 \sum \infty \frac{4}{5 n}

\frac{\partial}{\partial s} (\frac{( 3 r + s )}{t})