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)

\int x^{3} - x^{2} - 2 x d x

\int_{0}^{1} x e^{x^{2}} - e^{x} d x

\int_{0}^{π} cos^{4} (x) d x

\frac{y ^{^{'}}}{x} = 2 x^{2} - y

f (x) = - \frac{4 x}{( x - 2 ) ^{3}}

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