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

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Answer 1 · 2016-03-02T16:01:32

Method 1

Use the product and chain rules.

#d/dx(sin(2x)cos(2x)) = d/dx(sin(2x))cos(2x)+sin(2x)d/dx(cos(2x))#

# = [cos(2x)d/dx(2x)]cos(2x)+sin(2x)[-sin(2x)d/dx(2x)]#

# = 2cos^2(2x)-2sin^2(2x)#

You can use trigonometry to rewrite this.

Method 2

Use #sin(2theta) = 2sintheta cos theta# to write

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

Now use the chain rule

#d/dx (1/2sin(4x)) = 1/2 cos(4x)d/dx(4x)#

# = 1/2 cos(4x)*4 = 2cos(4x)#