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

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

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

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

- \frac{1}{2}

Solution steps

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

Treat

u

as a constant

Take the constant out:

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

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

Apply the Sum/Difference Rule:

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

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

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

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

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

Simplify

= - \frac{1}{2}

n = 0 \sum \infty \frac{n + 5}{5 ^{n}}

y^{'} = 0.1 (5 - y) y

\int \frac{3 ln ( x )}{x} d x

\frac{\partial}{\partial x} (- 2 y (- 2 x - y + 6))

\int_{- 8}^{0} \frac{1}{1 - x} d x

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