How do you differentiate f(x)=sin2x * cotxf(x)=sin2xcotx using the product rule?

1 Answer
Jan 24, 2016

2cos2x*cotx-csc^2x*sin2x2cos2xcotxcsc2xsin2x

Explanation:

  • The product rule:
    d/(dx)[f(x)]=("derivative of the first term" * "the second term")+("derivative of the second term"*"the first term")ddx[f(x)]=(derivative of the first termthe second term)+(derivative of the second termthe first term)

  • d/dx[cotx]=-csc^2x ddx[cotx]=csc2x

  • d/(dx)[f(x)]= (d/dx[sin(2x)]*cotx)+(d/dx[cotx]*sin2x)ddx[f(x)]=(ddx[sin(2x)]cotx)+(ddx[cotx]sin2x)
    =(2cos2x*cotx)+(-csc^2x*sin2x)=(2cos2xcotx)+(csc2xsin2x)
    =2cos2x*cotx-csc^2x*sin2x=2cos2xcotxcsc2xsin2x

  • You could just stop there, or you could simplify the answer further by uniting all the angles as xx using double angle formulas.