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

Lösung

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

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

Schritte zur Lösung

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

Behandle

x

als Konstante

Wende die Kettenregel an:

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

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

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

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

Vereinfache

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

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

\int cos (a ln (\frac{x}{b})) d x

t an g e n t f (x) = \frac{1}{x - 2}

\frac{d}{d x} (3 1 + tan (x))

y^{^{'}} - 2 x (y^{2} + 1) = 0

\frac{d}{d x} ((4 x + 1)^{2} (1 - x)^{3})