What is the derivative of 2cos(2t) ?

The derivative of 2cos(2t) is -4sin(2t)

What is the first derivative of 2cos(2t) ?

The first derivative of 2cos(2t) is -4sin(2t)

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derivative of

2 cos (2 t)

- 4 sin (2 t)

Solution steps

\frac{d}{d t} (2 cos (2 t))

Take the constant out:

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

= 2 \frac{d}{d t} (cos (2 t))

Apply the chain rule:

- sin (2 t) \frac{d}{d t} (2 t)

= - sin (2 t) \frac{d}{d t} (2 t)

\frac{d}{d t} (2 t) = 2

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

Simplify

= - 4 sin (2 t)

f (x) = \frac{x - 1}{x ^{2}}

\int \frac{1}{6 x lo g _{7} ( x )} d x

\int 3 x^{2} y + e^{y} d x

\int_{0}^{\frac{π}{2}} x cos (x) d x

\frac{d x}{d y} + 2 x y = 4 y

What is the derivative of 2cos(2t) ?
The derivative of 2cos(2t) is -4sin(2t)
What is the first derivative of 2cos(2t) ?
The first derivative of 2cos(2t) is -4sin(2t)