What is the derivative of #sin(2x)cos(2x)#?

2 Answers
Harish Chandra Rajpoot
Jul 21, 2018

#2\cos(4x)#

Explanation:

Given function:

#\sin (2x)\cos (2x)#

#1/2(2\sin (2x)\cos (2x))#

#1/2\sin (4x)#

Differentiating given function w.r.t. #x# as follows

#d/dx(1/2\sin(4x))#

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

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

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

#=2\cos(4x)#

Jim G.
Jul 21, 2018

#2cos4x#

Explanation:

#"differentiate using the "color(blue)"product/chain rule"#

#"given "y=f(x)g(x)" then"#

#dy/dx=f(x)g'(x)+g(x)f'(x)larrcolor(blue)"product rule"#

#f(x)=sin2xrArrf'(x)=2cos2x#

#g(x)=cos2xrArrg'(x)=-2sin2x#

#d/dx(sin2xcos2x)#

#=-2sin^2 2x+2cos^2 2x#

#=2(cos^2 2x-sin^2 2x)=2cos4x#

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