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

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Answer 1 · 2016-08-02T20:21:40

$f'(x) = sin(2x) + 2xcos(2x)$

We're going to be using the product rule anyway, but instead of having to deal with 3 terms we can simplify this by noticing that

$2sin(x)cos(x) = sin(2x)$

So $f(x)=xsin(2x)$

$f'(x) = d/(dx)(x)sin(2x) + xd/(dx)(sin(2x))$

First term, derivative of $x$ is $1$ .

We differentiate $sin(2x)$ using the chain rule.

$u = 2x implies (du)/(dx) = 2$

$y = sin(u) implies (dy)/(du) = cos(u)$

$implies (dy)/(dx) = 2cos(2x)$

$therefore f'(x) = sin(2x) + 2xcos(2x)$

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