This answer supposes #log(x)# refers to the natural (base-#e#) logarithm.*
Using the chain rule and the derivative of #log(x)#:
The chain rule, together with the known derivative #d/dxlog(x) = 1/x#, gives us
#d/dxlog(x-2) = 1/(x-2)(d/dx(x-2))#
#=1/(x-2)(1)#
#=1/(x-2)#
Using implicit differentiation and the derivative of #e^x#:
If we do not know the derivative of #log(x)#, but do know the derivative #d/dxe^x = e^x#, we can first create and modify an equation and use implicit differentiation:
Let #y = log(x-2)#
#=> e^y = e^log(x-2)#
#=> e^y = x-2#
#=> d/dxe^y = d/dx(x-2)#
#=> e^ydy/dx = 1#
#=> dy/dx = 1/e^y#
#=1/e^(log(x-2))#
#=1/(x-2)#
*If the intended logarithm is base-#10#, rather than the natural logarithm, then a coefficient of #1/ln(10)# should be included in the answer, as converting to the natural logarithm would give #log_10(x) = ln(x)/ln(10)#, and then the process would proceed as above.
