derivative of ln|coth(5x|)

Lösung

\frac{d}{d x} (ln ∣ coth (5 x) ∣)

- 5 csch^{2} (5 x) tanh (5 x)

Schritte zur Lösung

\frac{d}{d x} (ln ∣ coth (5 x) ∣)

Wende die Kettenregel an:

\frac{1}{∣ coth ( 5 x ) ∣} \frac{d}{d x} (∣ coth (5 x) ∣)

= \frac{1}{∣ coth ( 5 x ) ∣} \frac{d}{d x} (∣ coth (5 x) ∣)

\frac{d}{d x} (∣ coth (5 x) ∣) = - \frac{5 csch ^{2} ( 5 x ) coth ( 5 x )}{∣ coth ( 5 x ) ∣}

= \frac{1}{∣ coth ( 5 x ) ∣} (- \frac{5 csch ^{2} ( 5 x ) coth ( 5 x )}{∣ coth ( 5 x ) ∣})

Vereinfache

\frac{1}{∣ coth ( 5 x ) ∣} (- \frac{5 csch ^{2} ( 5 x ) coth ( 5 x )}{∣ coth ( 5 x ) ∣}) : - 5 csch^{2} (5 x) tanh (5 x)

= - 5 csch^{2} (5 x) tanh (5 x)

\frac{\partial}{\partial u ( v )} (\frac{e ^{v}}{u ( v ) + v ^{3}})

n = 2 \sum \infty \frac{( - 1 ) ^{n}}{n}

\int x^{3} e^{- 2 x} d x

\frac{d}{d x} (x^{2} + A x + B)

\int 4 x e^{4 x} d x