What is the derivative of y=sin(2x)cos(2x) ?

The derivative of y=sin(2x)cos(2x) is 2cos(4x)

What is the first derivative of y=sin(2x)cos(2x) ?

The first derivative of y=sin(2x)cos(2x) is 2cos(4x)

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derivative of

y = sin (2 x) cos (2 x)

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)

\frac{\partial}{\partial x} ((\frac{x + 2}{y + 1})^{\frac{1}{2}})

e^{a s}

n \to \infty lim (- 3 + 4 n^{5})

\frac{\partial}{\partial y} (8 z e^{x yz})

\frac{d}{d x} ((9 x^{2} + 8) (3 x + x))

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