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

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

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

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

- 2 sin (2 x) sin (2 t)

Solution steps

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

Treat

x

as a constant

Take the constant out:

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

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

Apply the chain rule:

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

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

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

= sin (2 x) (- sin (2 t) \cdot 2)

Simplify

= - 2 sin (2 x) sin (2 t)

\int (6 x^{2} + 2 x - 1) d x

\frac{( 2 s + 12 )}{s ^{2} + 2 s + 5}

\frac{\partial}{\partial x} ((x - 6) (y + 4))

\frac{d}{d x} (\frac{8 x}{7 - tan ( x )})

\frac{d y}{d t} = - 2 t y + 4 e^{- t^{2}}

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