#d/dx[(cosx-cos2x)/(1-cosx)]=#?

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Answer 1 · 2018-05-14T15:54:15

the answer
#[-((2*cos(x)-2)*sin(2x)-sin(x)*cos(2x)+sin(x))/(cos(x)-1)^2]#

show below

#d/dx[(cosx-cos2x)/(1-cosx)]=#

#[(1-cosx)*(-sinx+2sin2x)-(cosx-cos2x)(sinx)]/(1-cosx)^2#

#[-sinx+2sin2x+sinxcosx-2cosxsin2x-sinxcosx+sinxcos2x]/(1-cosx)^2#

#[-sinx+2sin2x-2cosxsin2x+sinxcos2x]/(1-cosx)^2#

#((1-cos(x))*(2*sin(2*x)-sin(x))-sin(x)*(cos(x)-cos(2*x)))/(1-cos(x))^2#

Rewrite/simplify

#(2*sin(2x)-sin(x))/(1-cos(x))-(sin(x)*(cos(x)-cos(2x)))/(1-cos(x))^2#

#-((2*cos(x)-2)*sin(2x)-sin(x)*cos(2x)+sin(x))/(cos(x)-1)^2#