You can also do it this way:
#log_x(10) = (log 10)/(log x) = 1/(log x)#
where #log x# is implied to be #log_10(x)#.
This works because #log_10(x)/(log_10(x)) = (k/k)(lnx)/(lnx)# where #k# is a constant. That's why you have a #log# and a #ln# formulation of the Nernst equation, for example:
#E = E^o - (RT)/(nF)lnQ#
#= E^o - (2.303RT)/(nF)logQ#
Since #lnx# and #logx# are constant when evaluated at a singular #x#, they differ by a constant factor. e.g.:
#log x = (lnx)/k => k = (lnx)/(logx) ~~ 2.302585cdots ~~ 2.303#
If taking the derivative of #f(x) = log_x(10)#, then because of the equality determined above:
#color(blue)(d/(dx)[(logx)^(-1)]) = -(logx)^(-2)*(d/(dx))[(lnx)/(ln10)]#
(Power Rule + Chain Rule)
The change of base law was also used, because I personally don't always remember the derivative of #log_10(x)# right away, but remembering it for #lnx# is somewhat easier.
# = -(logx)^(-2)*1/(ln10)1/x#
(#ln 10# is constant, and #d/(dx)[lnx] = 1/x#)
#= -1/((xln10)(log_10(x))^2)#
With the change of base law, you can adjust this even further to make it more uniform (one type of logarithm only):
#= -1/((xln10)((lnx)/(ln10))^2)#
#= -1/(xcancel(ln10))(ln10)^cancel(2)/(lnx)^2#
#color(blue)(= -(ln10)/(xln^2x))#
