What is (\partial)/(\partial x)(2cos(4x)) ?

The answer to (\partial)/(\partial x)(2cos(4x)) is -8sin(4x)

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(\partial)/(\partial x)(2cos(4x))

\frac{\partial}{\partial x} (2 cos (4 x))

- 8 sin (4 x)

Solution steps

\frac{\partial}{\partial x} (2 cos (4 x))

Take the constant out:

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

= 2 \frac{\partial}{\partial x} (cos (4 x))

Apply the chain rule:

- sin (4 x) \frac{\partial}{\partial x} (4 x)

= - sin (4 x) \frac{\partial}{\partial x} (4 x)

\frac{\partial}{\partial x} (4 x) = 4

= 2 (- sin (4 x) \cdot 4)

Simplify

= - 8 sin (4 x)

e^{e^{t}}

\int 5 sec^{2} (5 x) tan^{5} (5 x) d x

\frac{d}{d x} (\frac{( x + 1 )}{2})

\frac{d}{d t} (a (1 - cos (t)))

\frac{d y}{d t} = e^{t} (y - 1)^{2}

What is (\partial)/(\partial x)(2cos(4x)) ?
The answer to (\partial)/(\partial x)(2cos(4x)) is -8sin(4x)