How do you differentiate #f(x) =sin(3x-2)* cos(3x -2) #?

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Answer 1 · 2016-04-04T09:10:24

Remember that

#sin2A=2*sinA*cosA#

Hence

#f(x)=sin(3x-2)*cos(3x-2)=>f(x)=1/2*sin(2*(3x-2))#

Hence

#(df)/dx=1/2*cos(2(3x-2))*6=3*cos(6x-4)=#

Finally

#f'(x)=3*cos(6x-4)#