derivative of ln((e^x-1/(e^{x+1)}))

Lösung

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

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

Schritte zur Lösung

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

Vereinfache

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

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

Wende die Summen-/Differenzregel an:

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

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

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

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

\frac{d}{d x} (1) = 0

= \frac{e ^{x}}{e ^{x} - 1} - 1 - 0

Vereinfache

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

\int_{0}^{2} (x - 7)^{2} d x

\int 3 a x d x

x \to 3 - lim (\frac{1}{x ^{2} - 9})

x \to 2 lim (\frac{4 - x ^{2}}{x})

t an g e n t y = x + x, (16, 20)