How do you differentiate #g(x) = sqrt(x^2-x)sin2x# using the product rule?

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Answer 1 · 2018-06-05T07:51:50

#g'(x)=(4(x^2-x)cos2x+(2x-1)sin2x)/(2sqrt(x^2-x))#

#"We know the "color(blue)"Product Rule :"#

#color(blue)(y=u*v=>(dy)/(dx)=u*(dv)/(dx)+v*(du)/(dx)#

Let, #u=sqrt(x^2-x) and v=sin2x#

#:.(du)/(dx)=1/(2sqrt(x^2-x))(2x-1)=(2x-1)/(2sqrt(x^2-x)#

#and (dv)/(dx)=2cos2x#

So,

#(dy)/(dx)=sqrt(x^2-x)*2cos2x+sin2x(2x-1)/(2sqrt(x^2-x))#

#g'(x)=(2(x^2-x)*2cos2x+(2x-1)sin2x)/(2sqrt(x^2-x)#

#g'(x)=(4(x^2-x)cos2x+(2x-1)sin2x)/(2sqrt(x^2-x))#