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.
