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

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Answer 1 · 2016-01-09T06:40:43

The derivative is #2sin(x)cos(x)# or it can also be written as #sin(2x)#

#y=sin^2(x)#

Let us use the chain rule.

Let #y=u^2# and #u = sin(x)#

Chain rule:

#dy/dx = dy/(du)xx(du)/dx#

#y=u^2#

#dy/(du) = 2u#

#u=sin(x)#

#(du)/dx = cos(x)#

#dy/dx = dy/(du)xx(du)/dx#

#dy/dx = 2uxxcos(x)#

#dy/dx = 2sin(x)cos(x)#

#dy/dx = sin(2x)#