What is the d/(dt)(4sin(2t)) ?

The d/(dt)(4sin(2t)) is 8cos(2t)

What is the first d/(dt)(4sin(2t)) ?

The first d/(dt)(4sin(2t)) is 8cos(2t)

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\frac{d}{d t} (4 sin (2 t))

8 cos (2 t)

Solution steps

\frac{d}{d t} (4 sin (2 t))

Take the constant out:

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

= 4 \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

= 4 cos (2 t) \cdot 2

Simplify

= 8 cos (2 t)

\int \frac{3 x}{( x ^{2} + 5 ) ^{2}} d x

\frac{d}{d x} (\frac{11 - 11 x}{7 x ^{\frac{3}{7}}})

\frac{\partial}{\partial v} (\frac{1}{v})

\frac{d}{d x} (3 x^{2} - 12 x + 5)

y \cdot d x + x^{2} \cdot d y = 0