Single Variable Calculus, Chapter 7, Review Exercises, Section Review Exercises, Problem 58

Determine the equation of the tangent line to the curve $y = x \ln x$ at point $(e, e)$

Solving for the slope


$
\begin{equation}
\begin{aligned}

y' =& \frac{d}{dx} (x \ln x)
\\
\\
y' =& x \frac{d}{dx} (\ln x) + \ln x \frac{d}{dx} (x)
\\
\\
y' =& \cancel{x} \cdot \frac{1}{\cancel{x}} + \ln x
\\
\\
y' =& 1 + \ln x

\end{aligned}
\end{equation}
$


We know that $y' = m( \text{slope})$, so


$
\begin{equation}
\begin{aligned}

m =& 1 + \ln x
\\
\\
m =& 1 + \ln e
\\
\\
m =& 1 + 1
\\
\\
m =& 2

\end{aligned}
\end{equation}
$


Using Point Slope Form


$
\begin{equation}
\begin{aligned}

y - y_1 =& m (x - x_1)
\\
\\
y - e =& 2 (x - e)
\\
\\
y - e =& 2x - 2e
\\
\\
y =& 2x - 2e + e
\\
\\
y =& 2x - e \qquad \text{ Equation of the tangent line at } (e, e)

\end{aligned}
\end{equation}
$