y = (1+ log(x-numerator)3-base) / x
I didnt know how to write the log but the numerator is x and the base is 3.
Thanks in advance
$\displaystyle
f(x)=\frac{1+\log_3{x}}{x}
$
First lets use the change of base property to get rid of the $\displaystyle \log_3{x}$
$\displaystyle \log_3{x}=\frac{\ln{x}}{\ln{3}}$
Substituting this we get,
$\displaystyle f(x)=\frac{1+\frac{\ln{x}}{\ln{3}}}{x}$
Now to diferentiate this you need to either bring up the x and use the product rule or use the quotient rule.
$\displaystyle \frac{d}{dx}f(x)=\left(1+ \frac{1}{\ln{3}}\ln{x}\right)(-x^{-2})+(x^{-1})\left( \frac{1}{\ln{x}}\left(\frac{1}{x}\right)\right)$
Doing several steps of simplification and using a sum of logarithms property we get...
$\displaystyle \frac{d}{dx}f(x)=\frac{\ln{3x}-1}{x^2\ln{3}}$