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

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

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Answer 1 · 2018-07-21T18:09:10

$2 cos (4 x)$ $2\cos(4x)$

Explanation:

Given function:

$sin (2 x) cos (2 x)$ $\sin (2x)\cos (2x)$

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

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

Differentiating given function w.r.t. $x$ $x$ as follows

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

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

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

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

$= 2 cos (4 x)$ $=2\cos(4x)$

Answer 2 · 2018-07-21T18:38:43

$2 cos 4 x$ $2cos4x$

Explanation:

$differentiate using the product/chain rule$ $"differentiate using the "color(blue)"product/chain rule"$

$given y = f (x) g (x) then$ $"given "y=f(x)g(x)" then"$

$dy/dx=f(x)g'(x)+g(x)f'(x)larrcolor(blue)"product rule"$

$f(x)=sin2xrArrf'(x)=2cos2x$

$g(x)=cos2xrArrg'(x)=-2sin2x$

$d/dx(sin2xcos2x)$

$=-2sin^2 2x+2cos^2 2x$

$=2(cos^2 2x-sin^2 2x)=2cos4x$

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

2 Answers

Explanation:

Explanation:

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