How do you differentiate #sin^2(2x)#?

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Jim G. Share
Jul 23, 2016

Answer:

#2sin4x#

Explanation:

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

#color(red)(|bar(ul(color(white)(a/a)color(black)(d/dx(f(g(x)))=f'(g(x))g'(x))color(white)(a/a)|))) ........ (A)#

here #f(g(x))=sin^2(2x)=(sin2x)^2#

#rArrf'(g(x))=2sin2x#

Note, 2 applications of #color(blue)"chain rule"# required for g'(x)

#g(x)=sin2xrArrg'(x)=cos2x.d/dx(2x)=2cos2x#
#"------------------------------------------------------------------"#
Substitute these values into (A)

#2sin2x.2cos2x=4sin2xcos2x#

#color(orange)"Reminder" color(red)(|bar(ul(color(white)(a/a)color(black)(sin4x=2sin2xcos2x)color(white)(a/a)|)))#

#rArrd/dx(sin^2(2x))=4sin2xcos2x=2sin4x#

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