What is (\partial)/(\partial x)(-2sin(2x)) ?

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

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(\partial)/(\partial x)(-2sin(2x))

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

- 4 cos (2 x)

Solution steps

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

Take the constant out:

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

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

Apply the chain rule:

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

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

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

= - 2 cos (2 x) \cdot 2

Simplify

= - 4 cos (2 x)

arcsec (3 x)

y^{^{'}} = x^{2} y

4 t h \frac{1}{x - 5}

\frac{8 x ^{2} + 4 x + 2}{x}

f (x) = - 3 x x + 1

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