How do you find the differential #dy# of the function #y=1/3cos((6pix-1)/2)#?

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Answer 1 · 2017-02-01T22:21:55

#=- pi sin(3pix-1/2) #

#dy= d( 1/3cos((6pix-1)/2))#

factoring out the constant...
#=1/3 d( cos((6pix-1)/2))#

by the chain rule...
#=1/3 * -sin((6pix-1)/2) d( (6pix-1)/2)#

executing the chain rule...
#=- 1/3 sin((6pix-1)/2) (6pi)/2#

re-arranging the algebra...
#=- pi sin((6pix-1)/2) #

#=- pi sin(3pix-1/2) #

and as sine is an odd function....
#=pi sin(1/2 - 3pix) #