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

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

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

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

- \frac{1}{2}

Decimal

- 0.5

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}

s l o p e (- 3, - 8), (5, 6)

\frac{\partial}{\partial x} (x^{2} y + 2 y^{2} - 6 x + 3 y - 4)

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

\frac{d}{d y} (e^{y z})

\int_{1}^{2} \frac{6}{x ^{2}} d x

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