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)

x \to 0 lim (4 x + 48)

- 5 (t - 4)

\frac{d}{d x} (4 x^{2 π})

3 x^{2} - 2 x + 5

\frac{d y}{d x} = 8 x y

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