What is the d/(dt)(-8sin(t)) ?

The d/(dt)(-8sin(t)) is -8cos(t)

What is the first d/(dt)(-8sin(t)) ?

The first d/(dt)(-8sin(t)) is -8cos(t)

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\frac{d}{d t} (- 8 sin (t))

- 8 cos (t)

Solution steps

\frac{d}{d t} (- 8 sin (t))

Take the constant out:

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

= - 8 \frac{d}{d t} (sin (t))

Apply the common derivative:

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

= - 8 cos (t)

- \frac{5}{x ^{4}}

x \to - 1 lim (x^{2} - 3 x)

\int \frac{4 x}{1} d x

\frac{\partial}{\partial z} (y + z)

f (x) = 2 x + x