How do you find the derivative of #1/(sec x - tan x)#?

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Answer 1 · 2016-07-30T16:48:17

#1/(1-sinx)#

First, simplify the function:

#y=1/(1/cosx-sinx/cosx)=1/((1-sinx)/cosx)=cosx/(1-sinx)#

From here, use the quotient rule.

#y^'=(d/dx(cosx)*(1-sinx)-(cosx)*d/dx(1-sinx))/(1-sinx)^2#

#y^'=((-sinx)(1-sinx)-cosx(-cosx))/(1-sinx)^2#

#y^'=(-sinx+sin^2x+cos^2x)/(1-sinx)^2#

Recall that #sin^2x+cos^2x=1#:

#y^'=(1-sinx)/(1-sinx)^2#

#y^'=1/(1-sinx)#

Note that there are many different ways this can be written, but this is the simplest.