h(x) = Log(x) /x ------where "Log" is to the base 10.
Notes:
-----d/dx Log(base b)[u] = [1 /ln(b)]*(1/u) du/dx
-----To change log(base b)[A] to log(base c), then:
log(base b)[A] = log(base c)[A] / log(base c)[b]
So,
h'(x) = {x[(1/ln(10) *1/x] -Log(x)} /x^2
h'(x) = {1/ln(10) -Log(x)} /x^2
Convert the ln(10) to Log(10),
h'(x) = {[1 / Log(10)/Log(e)] -Log(x)} /x^2
h'(x) = {(Log(e) /Log(10)) -Log(x)} /x^2
h'(x) = {(Log(e) /1) -Log(x)} /x^2
h'(x) = {Log(e) -Log(x)} /x^2
h'(x) = Log(e/x) /x^2 ---------------answer.