What is the d/(dt)(-2sin(2t)) ?

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

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

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

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

- 4 cos (2 t)

Solution steps

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

Take the constant out:

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

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

Apply the chain rule:

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

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

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

= - 2 cos (2 t) \cdot 2

Simplify

= - 4 cos (2 t)

x \to 4 lim (4 x)

\int \frac{ln ( x )}{x x} d x

n = 1 \sum \infty \frac{1}{( n ( n + 2 ) )}

\int (\frac{1}{x ^{- 0.1}}) d x

y^{^{'}} = (x + y - 1)^{2}