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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply the Sum/Difference Rule:

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

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

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

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

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

Simplify

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

y = (\frac{( x + 2 )}{( x - 2 ) ^{2}})

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

y^{^{''}} + 6 y^{^{'}} + 9 = e^{4 t}, y (0) = 0, y^{^{'}} (0) = 0

\frac{d y}{d x} = \frac{1 - y ^{2}}{x}

\int \frac{18 - 12 x}{( 4 x - 1 ) ( x - 4 )} d x

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