Question

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

Answer 1

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

Explanation:

`"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))`