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

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

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(\partial)/(\partial x)(ln(e^x+e^y))

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

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

Solution steps

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

Treat

y

as a constant

Apply the chain rule:

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

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

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

= \frac{1}{e ^{x} + e ^{y}} e^{x}

Simplify

\frac{1}{e ^{x} + e ^{y}} e^{x} : \frac{e ^{x}}{e ^{x} + e ^{y}}

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

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

\int x tan (π x) d x

\frac{\partial}{\partial x} (sin (4 x - 2 y))

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

\frac{\partial}{\partial x} (x e^{- 3 x y})

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