What is the derivative of #(12/sinx) + (1/cotx)#?

1 Answer
Sep 16, 2015

#−12cosecx⋅cotx+sec^2x#

Explanation:

1/#sin x# = #cosec x# similarly #1/cotx# = #tanx#
the question changes to #12cosecx+tanx#
the derivative is #-12cosecx*cotx+sec^2x# by (u/v rule of differentiation)
#d/dx# #(1/sinx)# = #1/sin^2x# #(-1*cosx)#
= #-cosx/sinx# #1/sinx#
=#-cosecx cotx#
#d/dx# #tanx#= #d/dx(sinx/cosx)#
=#1/cos^2x# #(cosx*cosx-sinx*(-sinx))#
=#(cos^2x+sin^2x)/cos^2x#
=#1/cos^2x#=#sec^2x#