What is the d/(dt)(2cos(2t)) ?

The d/(dt)(2cos(2t)) is -4sin(2t)

What is the first d/(dt)(2cos(2t)) ?

The first d/(dt)(2cos(2t)) is -4sin(2t)

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\frac{d}{d t} (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)

\frac{d}{d x} (\frac{x}{5 x + 8})

x \to - 2 + lim (\frac{1 - x}{x + 2})

\frac{\partial}{\partial x} (x^{5})

\int cos (4 x) sin^{5} (4 x) d x

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