What is (\partial)/(\partial t)(2sin(t)) ?

The answer to (\partial)/(\partial t)(2sin(t)) is 2cos(t)

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(\partial)/(\partial t)(2sin(t))

\frac{\partial}{\partial t} (2 sin (t))

2 cos (t)

Solution steps

\frac{\partial}{\partial t} (2 sin (t))

Take the constant out:

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

= 2 \frac{\partial}{\partial t} (sin (t))

Apply the common derivative:

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

= 2 cos (t)

\int 6 sin^{5} (t) cos (t) d t

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

\frac{d}{d x} (a (a - x) (a - y) (a - z))

\int 6 x - 3 d x

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

What is (\partial)/(\partial t)(2sin(t)) ?
The answer to (\partial)/(\partial t)(2sin(t)) is 2cos(t)