What is (\partial)/(\partial u)(2cos(u)) ?

The answer to (\partial)/(\partial u)(2cos(u)) is -2sin(u)

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(\partial)/(\partial u)(2cos(u))

\frac{\partial}{\partial u} (2 cos (u))

- 2 sin (u)

Solution steps

\frac{\partial}{\partial u} (2 cos (u))

Take the constant out:

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

= 2 \frac{\partial}{\partial u} (cos (u))

Apply the common derivative:

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

= 2 (- sin (u))

Simplify

= - 2 sin (u)

y = - 2 x^{2} - 4 x + 1, (0, 1)

\frac{x}{2 x - 1}

\frac{d y}{d x} = x y + 3 x + 4 y + 12

\int ln (3 x - 2) d x

\int x^{3} + sin (x) d x

What is (\partial)/(\partial u)(2cos(u)) ?
The answer to (\partial)/(\partial u)(2cos(u)) is -2sin(u)