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Popular >

derivative of (sin(2x)(cos(2x)))

Solution

\frac{d}{d x} ((sin (2 x)) (cos (2 x)))

Solution

2 cos (4 x)

Solution steps

\frac{d}{d x} ((sin (2 x)) (cos (2 x)))

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = sin (2 x), g = cos (2 x)

= \frac{d}{d x} (sin (2 x)) cos (2 x) + \frac{d}{d x} (cos (2 x)) sin (2 x)

\frac{d}{d x} (sin (2 x)) = cos (2 x) \cdot 2

\frac{d}{d x} (cos (2 x)) = - sin (2 x) \cdot 2

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

Simplify

cos (2 x) \cdot 2 cos (2 x) + (- sin (2 x) \cdot 2) sin (2 x) : 2 cos (4 x)

= 2 cos (4 x)

Popular Examples

(\partial)/(\partial x)(x^{0.9})

\frac{\partial}{\partial x} (x^{0.9})

inverse oflaplace 1/(1-LS)inverse laplace

\frac{1}{1 - L S}

derivative of (-9/(x^2))

\frac{d}{d x} (\frac{- 9}{x ^{2}})

integral of e^{-2t}

\int e^{- 2 t} d t

y^'-8/x y=(y^4)/(x^{17)},y(1)=1

y^{^{'}} - \frac{8}{x} y = \frac{y ^{4}}{x ^{17}}, y (1) = 1

Frequently Asked Questions (FAQ)

What is the derivative of (sin(2x)(cos(2x))) ?
The derivative of (sin(2x)(cos(2x))) is 2cos(4x)
What is the first derivative of (sin(2x)(cos(2x))) ?
The first derivative of (sin(2x)(cos(2x))) is 2cos(4x)