(\partial)/(\partial x)(ln(e^x+e^{-x}))

Lösung

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

\frac{e ^{2 x} - 1}{e ^{2 x} + 1}

Schritte zur Lösung

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

Wende die Kettenregel an:

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

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

\frac{\partial}{\partial x} (e^{x} + e^{- x}) = e^{x} - e^{- x}

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

Vereinfache

\frac{1}{e ^{x} + e ^{- x}} (e^{x} - e^{- x}) : \frac{e ^{2 x} - 1}{e ^{2 x} + 1}

= \frac{e ^{2 x} - 1}{e ^{2 x} + 1}

\frac{\partial}{\partial x} (\frac{1}{a} e^{- (\frac{x}{a})^{2}})

t an g e n t f (x) = 8 x^{2} + 3 x, at x = - 45

\frac{d y}{d x} = 2 sec^{2} (x) e^{2 y}, y (0) = 0

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

\int \frac{1}{x + 4 + x + 3} d x