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How do you find the derivative of #(cosx)^x#?

1 Answer

Oct 5, 2015

#y'=(cosx)^x(ln(cosx)-xtgx)#

Explanation:

#y=(cosx)^x#
#lny=ln(cosx)^x#
#lny=xln(cosx)#

#d/dx(lny)=d/dx(xln(cosx))#

#1/y*y'=ln(cosx)+x*1/cosx*(-sinx)#

#(y')/y=ln(cosx)-xtgx#

#y'=y*(ln(cosx)-xtgx)#

#y'=(cosx)^x(ln(cosx)-xtgx)#

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