Respuesta :

f'(x) = [tex]e^{-x}[/tex] / x - [tex]e^{-x}[/tex]lnx

Differentiate using the product rule

given f(x) = g(x)h(x) then

f'(x) = g(x)h'(x) + h(x)g'(x) ← product rule

g(x) = [tex]e^{-x}[/tex] ⇒ g'(x) = - [tex]e^{-x}[/tex]

h(x) = lnx ⇒ h'(x) = [tex]\frac{1}{x}[/tex]

f'(x) = [tex]e^{-x}[/tex].[tex]\frac{1}{x}[/tex] - [tex]e^{-x}[/tex]lnx

      = [tex]e^{-x}[/tex] / x - [tex]e^{-x}[/tex]lnx