How do you differentiate $cos (3 x) sin (3 x)$ $cos(3x)sin(3x)$ ?

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Answer 1 · 2016-11-12T15:20:39

See below.

You could use the product rule and the chain rule, but I find it more satisfying to rewrite the function before differentiating.

From trigonometry:

$sin (2 θ) = 2 sin (θ) cos (θ)$ $sin(2theta) = 2sin(theta)cos(theta)$

Therefore

$sin (θ) cos (θ) = \frac{1}{2} sin (2 θ)$ $sin(theta)cos(theta) = 1/2sin(2theta)$

So,

$f (x) = cos (3 x) sin (3 x) = \frac{1}{2} sin (6 x)$ $f(x) = cos(3x)sin(3x) = 1/2sin(6x)$

Therefore,

$f'(x) = 1/2cos(6x) * d/dx(6x)$ $" "$ (chain rule)

$= 3cos(6x)$

How do you differentiate cos(3x)sin(3x)cos(3x)sin(3x)cos(3x)sin(3x)?