Răspuns:
[tex]\frac{\ln \left(x\right)}{\sqrt{x\left(\ln \left(x\right)-1\right)}}[/tex]
Explicație pas cu pas:
[tex]f' = \left(2\sqrt{x\left(\ln \left(x\right)-1\right)}\right)'\: = 2\left(\sqrt{x\left(\ln \left(x\right)-1\right)}\right)'[/tex]
Derivez [tex][\sqrt{x\left(\ln \left(x\right)-1\right)}]'[/tex] dupa formula [tex]\left(\sqrt{u}\right)'\:=\frac{u'}{2\sqrt{u}}[/tex] :
[tex]u = x\left(\ln \left(x\right)-1\right)\\ u'=x'\left(\ln \left(x\right)-1\right)+\left(\ln \left(x\right)-1\right)'\:x\\ u'=1\cdot \left(\ln \left(x\right)-1\right)+\frac{1}{x}x\\ u'=\ln \left(x\right)[/tex]
Deci:
[tex][\sqrt{x\left(\ln \left(x\right)-1\right)}]' = \frac{1}{2\sqrt{x\left(\ln \left(x\right)-1\right)}}\ln \left(x\right)[/tex]
Si obtinem:
[tex]f' = 2\cdot \frac{1}{2\sqrt{x\left(\ln \left(x\right)-1\right)}}\ln \left(x\right)[/tex]
Simplificam doiarii si obtinem:
[tex]f' = \frac{\ln \left(x\right)}{\sqrt{x\left(\ln \left(x\right)-1\right)}}[/tex]
