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

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

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d/(dt)(8sin(t))

\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)

y = \frac{- 4 x}{x ^{2} + 1}

5 x^{3} - 6 x^{2} + 12 x - 3, 3

\frac{d}{d x} (4 x^{3} + 3 x^{2} - 6 x + 1)

x \to 0 + lim (\frac{4}{x ^{2} - x})

\frac{\partial}{\partial y} (y e^{x y} + y)