How do you differentiate #g(t)=4sect+tant#?
1 Answer
May 2, 2017
Explanation:
The following trigonometric derivatives are very useful:
#d/dtsect=sect tant# #d/dttant=sec^2t#
Thus,
#g'(t)=4sect tant+sec^2t=sect(4tant+sect)#
We can derive both of these derivatives:
#sect=(cost)^-1#
#d/dxsect=-(cost)^-2d/dtcost=(-1)/cos^2t(-sint)#
#=sect tant#
And:
#tant=sint/cost#
#d/dttant=((d/dtsint)cost-sint(d/dtcost))/cos^2t=(cos^2t+sin^2t)/cos^2t#
#=sec^2t#
