Question

Intrebare de logartim derivat.

Avem in felul urmator (x lnx)'= (1+lnx)? Asa imi spune profesorul si nu inteleg de ce nu a derivat logaritmul natural. Eu asa am facut: (1+1/x)

Exercitiul complet este:
f'(x)=(x lnx)/(x^2+x+1)

formula e (F/G)'= (F'G-FG')/G^2

Stie careva?

Answer 1

[tex]\\F`_{(x)} = (\frac{xlnx}{x^2+x+1})` = \frac{\left(x\ln \left(x\right)\right)`\left(x^2+x+1\right)-\left(x^2+x+1\right)`x\ln \left(x\right)}{\left(x^2+x+1\right)^2} \\ \\ \\=\ \textgreater \ \frac{\left(\ln \left(x\right)+1\right)\left(x^2+x+1\right)-\left(2x+1\right)x\ln \left(x\right)}{\left(x^2+x+1\right)^2} = \frac{-x^2\ln \left(x\right)+x^2+x+\ln \left(x\right)+1}{\left(x^2+x+1\right)^2}[/tex]

[tex][x \:\ lnx]` = \mathrm{Aplicam}:\quad \left(f\cdot g\right)'=f'\cdot g+f\cdot g' \\=\ \textgreater \ f=x,\:g=\ln \left(x\right) \\=\ \textgreater \ =\frac{d}{dx}\left(x\right)\ln \left(x\right)+\frac{d}{dx}\left(\ln \left(x\right)\right)x = \boxed{1\cdot \ln \left(x\right)+\frac{1}{x}x}[/tex]

[tex] \boxed{Edit: \:\ Am \:\ atasat \:\ ex \:\ cerut \:\ in \:\ comentariu}[/tex]