What is (\partial}{\partial u}(\frac{v+u)/2) ?

The answer to (\partial}{\partial u}(\frac{v+u)/2) is 1/2

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(\partial}{\partial u}(\frac{v+u)/2)

\frac{\partial}{\partial u} (\frac{v + u}{2})

\frac{1}{2}

Solution steps

\frac{\partial}{\partial u} (\frac{v + u}{2})

Treat

v

as a constant

Take the constant out:

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

= \frac{1}{2} \frac{\partial}{\partial u} (v + u)

Apply the Sum/Difference Rule:

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

= \frac{1}{2} (\frac{\partial}{\partial u} (v) + \frac{\partial}{\partial u} (u))

\frac{\partial}{\partial u} (v) = 0

\frac{\partial}{\partial u} (u) = 1

= \frac{1}{2} (0 + 1)

Simplify

= \frac{1}{2}

\frac{d}{d x} (3 x^{3} y)

\int_{0}^{3} \frac{1}{x} d x

\int_{0}^{t} x e^{- x} d x

x y^{^{'}} - y = 2 x^{2} - 2 x - 2

x y^{^{'}} - y = x, y (1) = 10

What is (\partial}{\partial u}(\frac{v+u)/2) ?
The answer to (\partial}{\partial u}(\frac{v+u)/2) is 1/2