Posted by **Charles** on Sunday, September 28, 2008 at 5:11pm.

How is 2cos(4x) the derivative of f(x)=sin(2x)cos(2x)?

- Calculus -
**bobpursley**, Sunday, September 28, 2008 at 5:57pm
If you take the derivative of sin2xcos2x, you get a form of cos^2 2x + sin^2 2x, then you use the double angle formulas to get cos4x

- Calculus -
**drwls**, Sunday, September 28, 2008 at 6:19pm
You can also start with identity

f(x) = sin 2x cos 2x = (1/2) sin 4x

Then take the derivative with respect to x. Using the chain rule with u(x) = 4x,

df/dx = df/du du/dx

you get

df/dx = (1/2) cos (4x) *4 = 2 cos (4x)

- Calculus -
**Charles**, Sunday, September 28, 2008 at 7:44pm
drwls: thanks for your help. would you mind explaining how you used that identity to set up the new problem? i get the chain rule but i had no idea that sin2xcos2x was equal to 1/2sin4x. thanks again.

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