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)

5 t

\frac{d}{d x} (\frac{1}{x ^{2} ( 3 x + 1 )})

\frac{\partial}{\partial x} (x y^{2} cos (2 + x^{2} y))

2 sin (x), [0, π]

f (x) = \frac{5}{2 5 x + 6}, at x = 6

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