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

The first (2cos(2x))^' is -4sin(2x)

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(2cos(2x))^'

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

- 4 sin (2 x)

Solution steps

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

Take the constant out:

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

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

Apply the chain rule:

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

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

(2 x)^{'} = 2

= 2 (- sin (2 x) \cdot 2)

Simplify

= - 4 sin (2 x)

3 s + 2

x \to \infty lim (e^{x} - 3)

\frac{d}{d x} (\frac{z ( x )}{x})

\int_{0}^{2} (y - 2)^{2} d y

\frac{d}{d x} ((2 x + 1)^{2} (x^{2} + x + 1))