How do you find the fourth derivative of #sin x#?

Question

11939 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-04-19T14:36:23

#y=sinx#
#=>y_1=d/(dx)(sinx)#
#=>y_1=cosx#
#=>y_2=d/(dx)(cosx)#
#=>y_2=-sinx#
#=>y_3=d/(dx)(-sinx)#
#=>y_3=-cosx#
#=>y_4=d/(dx)(-cosx)#
#=>y_4=sinx#