Question Number 172190 by Mikenice last updated on 23/Jun/22
Commented by mokys last updated on 23/Jun/22
![ln(f(x))=cos(x^2 +4x+2) lnx ((f^′ (x))/(f(x))) = ((cos(x^2 +4x+2))/x) − (2x +4)sin(x^2 +4x+2)lnx f^′ (x) = f(x) [ ((cos(x^2 +4x+2)−(2x^2 +4x)lnx sin(x^2 +4x+2))/x)] f^′ (x) = X^(cos(x^2 +4x+2)) [ ((cos(x^2 +4x+2)−(2x^2 +4x)lnx sin(x^2 +4x+2))/x) ] Aldolimy](https://www.tinkutara.com/question/Q172194.png)
$${ln}\left({f}\left({x}\right)\right)={cos}\left({x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{2}\right)\:{lnx} \\ $$$$ \\ $$$$\frac{{f}\:^{'} \left({x}\right)}{{f}\left({x}\right)}\:=\:\frac{{cos}\left({x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{2}\right)}{{x}}\:−\:\left(\mathrm{2}{x}\:+\mathrm{4}\right){sin}\left({x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{2}\right){lnx} \\ $$$$ \\ $$$${f}\:^{'} \left({x}\right)\:=\:{f}\left({x}\right)\:\left[\:\frac{{cos}\left({x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{2}\right)−\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{4}{x}\right){lnx}\:{sin}\left({x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{2}\right)}{{x}}\right] \\ $$$$ \\ $$$${f}\:^{'} \left({x}\right)\:=\:{X}\:^{{cos}\left({x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{2}\right)} \:\left[\:\frac{{cos}\left({x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{2}\right)−\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{4}{x}\right){lnx}\:{sin}\left({x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{2}\right)}{{x}}\:\right] \\ $$$$ \\ $$$${Aldolimy} \\ $$