How do you find the derivative of $sin (2 x) cos (2 x)$ $sin(2x)cos(2x)$ ?

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How do you find the derivative of $sin (2 x) cos (2 x)$ $sin(2x)cos(2x)$ ?

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Answer 1 · 2016-03-02T16:01:32

Method 1

Use the product and chain rules.

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

$= [cos (2 x) \frac{d}{d x} (2 x)] cos (2 x) + sin (2 x) [- sin (2 x) \frac{d}{d x} (2 x)]$ $= [cos(2x)d/dx(2x)]cos(2x)+sin(2x)[-sin(2x)d/dx(2x)]$

$= 2 {cos}^{2} (2 x) - 2 {sin}^{2} (2 x)$ $= 2cos^2(2x)-2sin^2(2x)$

You can use trigonometry to rewrite this.

Method 2

Use $sin (2 θ) = 2 sin θ cos θ$ $sin(2theta) = 2sintheta cos theta$ to write

$sin (2 x) cos (2 x) = \frac{1}{2} sin (4 x)$ $sin(2x)cos(2x)=1/2sin(4x)$

Now use the chain rule

$\frac{d}{d x} (\frac{1}{2} sin (4 x)) = \frac{1}{2} cos (4 x) \frac{d}{d x} (4 x)$ $d/dx (1/2sin(4x)) = 1/2 cos(4x)d/dx(4x)$

$= \frac{1}{2} cos (4 x) \cdot 4 = 2 cos (4 x)$ $= 1/2 cos(4x)*4 = 2cos(4x)$

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How do you find the derivative of $sin (2 x) cos (2 x)$ $sin(2x)cos(2x)$ ?

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