If #f(x) =sin^3x # and #g(x) = sqrt(3x-1 #, what is #f'(g(x)) #?

1 Answer
Jim S
May 7, 2018

#f(x)=sin^3x# , #D_f=RR#

#g(x)=sqrt(3x-1)#, #Dg=[1/3,+oo)#

#D_(fog)={##AAx##in##RR:##x##in##D_g#, #g(x)##in##D_f}#

#x>=1/3# , #sqrt(3x-1)##in##RR# #-># #x##in##[1/3,+oo)#

#AAx##in##[1/3,+oo)#,

  • #(fog)'(x)=f'(g(x))g'(x)=f'(sqrt(3x-1))((3x-1)')/(2sqrt(3x-1))#

#f'(x)=3sin^2x(sinx)'=3sin^2xcosx#

so #(fog)'(x)=sin^2(sqrt(3x-1))cos(sqrt(3x-1))*9/(2sqrt(3x-1))#

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