What is the derivative of #f(x) = sin2x*cos2x#?

What is the derivative of #f(x) = sin2x*cos2x#?

1 Answer
Nov 22, 2017

#d/dx[f(x)]=-2[sin^2(2x)-cos^2(2x)]#

Explanation:

Some basic rules of differentiation are as follows, where #u# and #v# are functions of #x#:

  1. Addition / subtraction rule
    If #y=u+-v#, #dy/dx=d/dx(u)+-d/dx(v)#

  2. Chain rule
    #dy/dx=dy/(du)xx(du)/dx#

  3. Product rule
    If #y=uv#, #dy/dx=u(dv)/dx+v(du)/dx#

  4. Quotient rule
    If #y=u/v#, #dy/dx=(v(du)/dx-u(dv)/dx)/v^2#

Let's get started,

#f(x) = sin2x*cos2x#
#d/dx[f(x)]=(sin2x)[d/dx(cos2x)]+(cos2x)[d/dx(sin2x)]# ( Product rule )
#color(white)(d/dx[f(x)])=(sin2x)(-2)(sin2x)+(cos2x)(2)(cos2x)#
#color(white)(d/dx[f(x)])=-2[sin^2(2x)-cos^2(2x)]#