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

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

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(cos(2x))^'

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

- sin (2 x) \cdot 2

Solution steps

(cos (2 x))^{'}

Apply the chain rule:

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

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

(2 x)^{'} = 2

= - sin (2 x) \cdot 2

\frac{d}{d x} (2 \cdot x)

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

\frac{d}{d x} (6 e^{3 x})

\int (a^{2} - x^{\frac{2}{3}}) d x

x \to 0 + lim (ln (1 + sin (x)))