Single Variable Calculus, Chapter 7, 7.2-1, Section 7.2-1, Problem 48

Determine the equation of the tangent line to the curve $\displaystyle y = \frac{e^x}{x}$ at the point $(1, e)$.


$
\begin{equation}
\begin{aligned}

\text{if } y =& \frac{e^x}{x} , \text{ then by using Quotient Rule...}
\\
\\
y' =& \frac{x(e^x) - e^x(1) }{(x)^2}
\\
\\
y' =& \frac{e^x(x - 1)}{x^2}

\end{aligned}
\end{equation}
$


Recall that the first derivative is equal to the slope of the line at some point...

So when $x = 1$,

$\displaystyle y' = m = \frac{e^1 (1 - 1)}{(1^2)} = 0$

Therefore, the equation of the tangent line can be determined using point slope form


$
\begin{equation}
\begin{aligned}

y - y_1 =& m(x - x_1)
\\
\\
y - e =& 0 (x - 1)
\\
\\
y =& e

\end{aligned}
\end{equation}
$

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