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

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Answer 1 · 2018-04-18T03:36:03

#d/dxsin(x/2)=1/2cos(x/2)#

The Chain Rule, when applied to the sine, tells us that

#d/dxsin(u)=cosu*(du)/dx#, where #u# is some function in terms of #x.# Here, we see #u=x/2,# so

#d/dxsin(x/2)=cos(x/2)*d/dx(x/2)#

#d/dx(x/2)=1/2,# so we end up with

#d/dxsin(x/2)=1/2cos(x/2)#

Answer 2 · 2018-04-18T03:36:39

#cos(x/2)/2#

use chain rule:
#d/dx(sin(x/2))=cos(x/2)*d/dx(x/2)#

(note that derivative of #sinx# is #cosx#)

#cos(x/2)*(1/2)#
#=cos(x/2)/2#