How do you find the derivative of #sin^2(2x)#?

1 Answer
Feb 3, 2017

I got: #f'(x)=4sin(2x)cos(2x)#

Explanation:

Here I would try to use the Chain Rule applied to three functions one nested into the other:
the first and all-embracing function is #()^2#; the next one is the #sin# function and the last one the #2x# function.

I will use the Chain Rule deriving each one as if alone (regardless of the argument) and I will multiply each individual derivative together using, as visual help, a sequence of red-blue-green colors to identify each derivative:

#f'(x)=color(red)(2sin^(2-1)(2x))*color(blue)(cos(2x))*color(green)(2)#
giving:
#f'(x)=4sin(2x)cos(2x)#

This function can be compressed (giving: #2sin(4x)#) using a trigonometric identity but I do not want to confuse the procedure.