How do you differentiate #y = (sec x) (tan x)#?
1 Answer
Feb 6, 2016
Explanation:
Use the product rule:
#y'=tanxd/dx(secx)+secxd/dx(tanx)#
You should know the identities:
#d/dx(secx)=tanxsecx# #d/dx(tanx)=sec^2x#
Which gives...
#y'=tanx(tanxsecx)+secx(sec^2x)#
Simplified:
#y'=tan^2xsecx+sec^3x#
