So I have the problem
$y=(ln3)log_3x$
and one of the steps is
$y=(ln3)(1/(x * ln3))$
then the ln cancel out and I understand all of that. I'm just wondering why ln3 doesn't go to 1 if it is just a constant?
Thanks! Please fix the formatting if you can. I'm not sure how to do format division.
$$y=\ln(3)\log_x(3)=\ln(3)\frac{\ln(3)}{\ln(x)}$$ $$y=(\ln(3))^2(\ln(x))^{-1}$$
Derivate $y$ with respect to $x$ and use chain rule in $(\ln(x))^{-1}$. Then,
$$\dot y = \color{red}{(\ln(3))^2} \left(-(\ln(x))^{-2}x^{-1}\right)=-\color{red}{(\ln(3))^2}\frac{1}{x(\ln(x))^2}$$
$\color{red}{(\ln(3))^2}$: this multiplicative factor is a constant*, so you don't have to worry when you derive.
*Thanks, John Hughes