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

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

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

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

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

Solution steps

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

Treat

x

as a constant

Apply the chain rule:

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

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

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

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

Simplify

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

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

y^{^{''}} + 8 y^{^{'}} + 41 y = 0

arccos (1 - x^{2})

\frac{d}{d x} (\frac{( x - 1 )}{2})

\int \frac{1}{7 - 4 x - x ^{2}} d x

\frac{( s + 1 )}{( s ^{2} + 2 s + 9 )}

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