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

The answer to (\partial)/(\partial y)(ln(x+e^{-y})) is -(e^{-y})/(x+e^{-y)}

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

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

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

Solution steps

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

Treat

x

as a constant

Apply the chain rule:

\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})

Simplify

\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

f (x) = \frac{1}{x - 2}

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

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

What is (\partial)/(\partial y)(ln(x+e^{-y})) ?
The answer to (\partial)/(\partial y)(ln(x+e^{-y})) is -(e^{-y})/(x+e^{-y)}