#### Answer

The equation of the tangent line at the given point is $y=e$

#### Work Step by Step

$y=\dfrac{e^{x}}{x}$ $;$ $a=1$
First, evaluate the derivative of the given expression using the quotient rule:
$y'=\dfrac{(x)(e^{x})'-(x)'e^{x}}{x^{2}}=\dfrac{xe^{x}-e^{x}}{x^{2}}=\dfrac{e^{x}(x-1)}{x^{2}}$
Substitute $x$ by $a=1$ in the derivative found to obtain the slope of the tangent line at the given point:
$m_{\text{tan}}=\dfrac{e^{1}(1-1)}{1^{2}}=\dfrac{e(0)}{1}=0$
Substitute $x$ by $a=1$ in the original expression to obtain the $y$-coordinate of the point given:
$y=\dfrac{e^{1}}{1}=e$
The point is $(1,e)$
The slope of the tangent line and a point through which it passes are known. Use the point-slope form of the equation of a line, which is $y-y_{1}=m(x-x_{1})$ to obtain the equation of the tangent line at the given point:
$y-e=0(x-1)$
$y-e=0$
$y=e$
The graph of both the function and the tangent line are shown the answer section.