What is the derivative of #sin^2(x) * cos^2(x)#?

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Answer 1 · 2016-11-07T01:19:44

By the chain and product rules, #d/dx(sin^2(x)*cos^2(x)) =-2cos(x)sin(x)(sin^2(x)-cos^2(x))#

In order to evaluate this derivative, we need to use both the product and chain rules.

Starting with #sin^2(x)# and using the chain rule to take its derivative, we have:

#2sin(x)*cos(x)#

Now, by the product rule, we multiply this by our second term, #cos^2(x)#, so the left side of the derivative is

#2sin(x)*cos(x)*cos^2(x)# or #2cos^3(x)sin(x)#

Now for the right side, we use the chain rule to take the derivative of #cos^2(x)#:

#2cos(x)*(-sinx)# or #-2cos(x)sin(x)#

Similarly to with the left side, we now multiply this by our #sin^(x)# term, so the right side of the derivative is

#-2cos(x)sin(x)*sin^2(x)# or #-2sin^3(x)cos(x)#

Continuing with the product rule, we add the left- and right-hand derivatives we calculated above together, so our final answer is:

#2cos^3(x)sin(x)-2sin^3(x)cos(x))#

This can be simplified in several ways, but one simplified version of the derivative may be:

#-2cos(x)sin(x)(sin^2(x)-cos^2(x))#

Answer 2 · 2016-11-19T02:10:20

#d/dxsin^2(x)cos^2(x)=sin(2x)cos(2x)#

Another method, using the chain rule along with the trigonometric identity #sin(2x) = 2sin(x)cos(x)#

#sin^2(x)cos^2(x) = (2sin(x)cos(x))^2/4 = sin^2(2x)/4#

#=> d/dxsin^2(x)cos^2(x) = d/dxsin^2(2x)/4#

#=1/4d/dxsin^2(2x)#

#=1/4*2sin(2x)(d/dxsin(2x))#

#=sin(2x)/2*cos(2x)(d/dx2x)#

#=(sin(2x)cos(2x))/2*2#

#=sin(2x)cos(2x)#