How do you find the derivative of #(cos^2(x)sin^2(x))#?

1 Answer
Oct 17, 2016

Answer:

#sin2xcos2x#

Explanation:

In this exercise we have to apply : two properties
the derivative of product:
#color(red)((uv)'=u'(x)v(x)+v'(x)u(x))#

The derivative of a power:
#color(blue)((u^n(x))'=n(u)^(n-1)(x)u'(x))#

In this exercise let :
#color(brown)(u(x)=cos^2(x))#
#color(blue)(u'(x)=2cosxcos'x)#
#u'(x)=-2cosxsinx#

Knowing the trigonometric identity that says :
#color(green)(sin2x=2sinxcosx)#

#u'(x)=-color(green)(sin2x)#

Let:
#color(brown)(v(x)=sin^2(x))#
#color(blue)(v'(x)=2sinxsin'x)#
#v'(x)=2sinxcosx#
#v'(x)=color(green)(sin2x)#

So,
#(cos^2xsin^2x)'#
#=color(red)((uv)'#
#=color(red)(u'(x)v(x)+v'(x)u(x))#
#=(-sin2x)(sin^2x)+sin(2x)cos^2x#
#=sin2x(cos^2x-sin^2x)#
Knowing the trigonometric identity that says :
#color(green)(cos2x=cos^2x-sin^2x)#

Therefore,
#(cos^2xsin^2x)'=sin2xcos2x#