Find the Derivative:
$\displaystyle h(x) = \frac{\log_{10}(x)}{x}$
the answer is $\displaystyle \frac{1}{x^2}(\log_{10}\frac{e}{x})$
How??
You need the change of base formula for logarithms:
$\displaystyle \log_a(b)=\frac{\log_c(b)}{\log_c(a)}$,
or in this case:
$\displaystyle \log_{10}(x)=\frac{\log_e(x)}{\log_e(10)}$.
So:
$\displaystyle h(x) = \frac{\log_{e}(x)}{x\,\log_e(10)}$.
This may be differentiated using the quotiont rule (or the product rule):
$\displaystyle \frac{dh}{dx}=\frac{1}{\log_e(10)}\left( -\frac{\log_e(x)}{x^2}+\frac{1}{x^2}\right)$
......$\displaystyle =\frac{1}{\log_e(10)}\left( -\frac{\log_e(x)}{x^2}+\frac{\log_e(e)}{x^2}\right) =\frac{1}{x^2}\left(\frac{ \log_{e}(e/x)}{\log_e(10)} \right)==\frac{1}{x^2}\left({ \log_{10}(e/x)} \right)$
RonL
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.