What is (\partial)/(\partial v)(4ucos(v)) ?

The answer to (\partial)/(\partial v)(4ucos(v)) is -4usin(v)

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(\partial)/(\partial v)(4ucos(v))

\frac{\partial}{\partial v} (4 u cos (v))

- 4 u sin (v)

Solution steps

\frac{\partial}{\partial v} (4 u cos (v))

Treat

u

as a constant

Take the constant out:

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

= 4 u \frac{\partial}{\partial v} (cos (v))

Apply the common derivative:

\frac{\partial}{\partial v} (cos (v)) = - sin (v)

= 4 u (- sin (v))

Simplify

= - 4 u sin (v)

\frac{\partial}{\partial x} (\frac{1}{1 + e ^{- x}})

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

\int \frac{- x}{1 + x} d x

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

\int_{0}^{2} x d x

What is (\partial)/(\partial v)(4ucos(v)) ?
The answer to (\partial)/(\partial v)(4ucos(v)) is -4usin(v)