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What is the derivative of # (cos x) / (1-sinx)#?

1 Answer

Jun 6, 2015

Use the Quotient Rule:
#y'=(-sinx(1-sinx)+cosx(cosx))/(1-sinx)^2=#
#=(-sinx+sin^2x+cos^2x)/(1-sinx)^2=#
#=(1-sinx)/(1-sinx)^2=1/(1-sinx)#

I used the fact that: #sin^2x+cos^2x=1#

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