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

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

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

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

- 4 sin (2 x)

Solution steps

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

Take the constant out:

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

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

Apply the chain rule:

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

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

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

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

Simplify

= - 4 sin (2 x)

\int \frac{5 x + 3}{x ^{2} + 4 x + 10} d x

\frac{d}{d x} (cos^{3} (\frac{x}{3}))

y^{^{''}} - 4 y^{^{'}} + 21 y = 0

\int (x^{2} + sin (x)) d x

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

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