How do you find the derivative of #h(x)= sin2xcos2x#?
1 Answer
Jun 6, 2015
We must use the product rule here, which states that, be
Thus, let's just bear in mind the derivatives of these trigonometric functions:
#y=sin(u)# , then#y'=u'cosu# #y=cos(u)# , then#y'=-u'sinu#