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

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

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(2cos(x))^'

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

- 2 sin (x)

Solution steps

(2 cos (x))^{'}

Take the constant out:

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

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

Apply the common derivative:

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

= 2 (- sin (x))

Simplify

= - 2 sin (x)

y^{2} = x - 2, y = x - 4

\int (sec (2 x) + tan (2 x)) d x

f (x) = \frac{2 t + 1}{t + 3}

y = 2 x^{2} + 3, (0, 3)

\int \frac{x ^{2} + 1}{1 - x ^{2}} d x