Derivade of: Y=e^(-x)Lnx

To find the derivative of y =
(e^-x)*Lnx.

y = e^(-x)/(
Lnx)

We take logarithms of both
sides:

logy =
Ln(e^-x)+Ln(Ln(x))

Lny = -x
+Ln(Ln(x))

We differentiate both sides with respect to
x:

(1/y)(dy/dx) =
{-x+Ln(ln(x)}'.

(1/y)(dy/dx) = (-x)'+{ Ln(Ln(x)}'
.

(1/y)(dy/dx) = -1+{1/(Ln(x)}(Ln(x)'
.

 (1/y)(dy/dx) = -1+1/(Ln(x))^2 , as (i) {Ln(x)} = 
1/Ln(x) and d/dx{u(v(x)} = (du/dv)(dv/dx).

dy/dx = y{-1
+1/(Ln(x))^2}.

dy/dx = {-1+1/(Ln(x))^2}
(e^-x)(Ln(x)).