How do you differentiate f(x)=sin2x * cotxf(x)=sin2x⋅cotx using the product rule?
1 Answer
Jan 24, 2016
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 term⋅the second term)+(derivative of the second term⋅the 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)=(2cos2x⋅cotx)+(−csc2x⋅sin2x)
=2cos2x*cotx-csc^2x*sin2x=2cos2x⋅cotx−csc2x⋅sin2x -
You could just stop there, or you could simplify the answer further by uniting all the angles as
xx using double angle formulas.