What is the derivative of #y=log_x 3#?

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Answer 1 · 2015-10-18T22:37:25

#y'=-1/(xln3(log_3x)^2)#

#y=log_x3=1/log_3x = (log_3x)^-1#

#y'=-(log_3x)^-2*1/(xln3)#

#y'=-1/(xln3(log_3x)^2)#

Answer 2 · 2015-11-16T02:09:54

#y'=-ln3/(x(lnx)^2#

#log_x3=ln3/lnx# through the change of base formula.
#d/(dx)[ln3/lnx]=ln3d/(dx)[1/lnx]=ln3color(blue)(d/(dx)[(lnx)^-1]#

About to use the Chain Rule:
#color(blue)(d/(dx)[(lnx)^-1])=-(lnx)^-2d/(dx)[lnx]=-1/(lnx)^2(1/x)#
#y'=-ln3/(x(lnx)^2#