Respuesta :

[tex](e^{-x})'=e^{-x}\cdot(-x)'=e^{-x}\cdot(-1)=-e^{-x}[/tex]
y = [tex] e^{-x}[/tex]                

let u = x

y = [tex] e^{u} [/tex]

dy/dx = dy/du  * du/dx

dy/du =  [tex] e^{u} [/tex]

du/dx = -1

dy/dx =  [tex] e^{u} [/tex]  * -1

dy/dx =  -[tex] e^{u} [/tex]

dy/dx = -[tex] e^{-x}[/tex]