Answer
$y=\frac{x}{e}-\frac{1}{e}$
Work Step by Step
$f(x)=e^{-x}ln(x)$ $(1,0)$
$u=e^{-x}$
$u’=-e^{-x}$
$v=ln(x)$
$v’=\frac{1}{x}$
$f’(x)=(-e^{-x})(ln(x))+(e^{-x})(\frac{1}{x})$
$=e^{-x}(-ln(x)+\frac{1}{x})$
$f’(1)=M=e^{-1}(-ln(1)+1)$
$=\frac{1}{e}$
Equation of Tangent: $y-0=\frac{1}{e}(x-1)$
$y=\frac{x}{e}-\frac{1}{e}$