What is the first (2sin(x))^' ?

The first (2sin(x))^' is 2cos(x)

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(2sin(x))^'

(2 sin (x))^{^{'}}

2 cos (x)

Solution steps

(2 sin (x))^{'}

Take the constant out:

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

= 2 (sin (x))^{^{'}}

Apply the common derivative:

(sin (x))^{'} = cos (x)

= 2 cos (x)

(- 29, - 40), (- 28, 3)

x \to - \infty lim (x^{- 5})

\int (- 3 x^{- 4}) d x

\frac{\partial}{\partial x} (\frac{1}{\frac{1}{x} + \frac{1}{y} + \frac{1}{z}})

\frac{d}{d x} (6 x - 12)