What is the derivative of #tanx/sinx#?

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Answer 1 · 2018-05-03T18:11:38

#d/dx{tanx/sinx}=d/dx{sinx/cosx*1/sinx}=d/dx{1/cosx}#
#=d/dx{secx}=secxtanx#.

Answer 2 · 2018-05-04T12:49:06

#secxtanx#

Given: #d/dx(tanx/sinx)#.

Notice how #tanx=sinx/cosx#.

#=d/dx(sinx/cosx*1/sinx)#

#=d/dx(1/cosx)#

Now, we know that #1/cosx=secx#.

#=d/dx(secx)#

This is a well known integral, where it equals,

#=secxtanx#