What is (\partial}{\partial y}(\frac{e^x+e^y)/2) ?

The answer to (\partial}{\partial y}(\frac{e^x+e^y)/2) is 1/2 e^y

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(\partial}{\partial y}(\frac{e^x+e^y)/2)

\frac{\partial}{\partial y} (\frac{e ^{x} + e ^{y}}{2})

\frac{1}{2} e^{y}

Solution steps

\frac{\partial}{\partial y} (\frac{e ^{x} + e ^{y}}{2})

Treat

x

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= \frac{1}{2} \frac{\partial}{\partial y} (e^{x} + e^{y})

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{1}{2} (\frac{\partial}{\partial y} (e^{x}) + \frac{\partial}{\partial y} (e^{y}))

\frac{\partial}{\partial y} (e^{x}) = 0

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

= \frac{1}{2} (0 + e^{y})

Simplify

= \frac{1}{2} e^{y}

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

x \to - 5 + lim (\frac{( x - 5 )}{x + 5})

\frac{d}{d x} ((6 x + 12)^{3})

\int \frac{x ^{2} + 2 x + 2}{x + 2} d x

x \to - 2 + lim (\frac{- 8 x}{x + 2})

What is (\partial}{\partial y}(\frac{e^x+e^y)/2) ?
The answer to (\partial}{\partial y}(\frac{e^x+e^y)/2) is 1/2 e^y