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

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

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

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

\frac{1}{2}

Solution steps

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

Treat

v

as a constant

Take the constant out:

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

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

Apply the Sum/Difference Rule:

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

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

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

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

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

Simplify

= \frac{1}{2}

\frac{\partial}{\partial x} (x^{2} y cos (\frac{z}{t}))

\frac{\partial}{\partial x} (\frac{- 3 x - y}{y ^{2} + 2 x ^{2}})

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

y^{^{'}} = 20 x - 2

\frac{d}{d x} (2 x + e^{x} + e^{- x})

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