Single Variable Calculus, Chapter 3, 3.5, Section 3.5, Problem 54

Determine the equation of the tangent line to the curve $y= \sqrt{5 + x^2}$ at the point $(2, 3)$.

Solving for the slope


$
\begin{equation}
\begin{aligned}

y' = m =& \frac{d}{dx} (\sqrt{5 + x^2})
\\
\\
m =& \frac{d}{dx} (5 + x^2)^{\frac{1}{2}}
\\
\\
m =& \frac{1}{2} (5 + x^2) ^{\frac{-1}{2}} \cdot \frac{d}{dx} (5 + x^2)
\\
\\
m =& \frac{1}{2} (5 + x^2) ^{\frac{-1}{2}} (0 + 2x)
\\
\\
m =& \frac{1}{\cancel{2}} (5 + x^2) ^{\frac{-1}{2}} (\cancel{2}x)
\\
\\
m =& x (5 + x^2) ^{\frac{-1}{2}}
\\
\\
m =& 2[5 + (2)^2] ^{\frac{-1}{2}}
\\
\\
m =& 2 [ 5 + 4 ] ^{\frac{-1}{2}}
\\
\\
m =& \frac{2}{\sqrt{9}}
\\
\\
m =& \frac{2}{3}


\end{aligned}
\end{equation}
$



Using the Point Slope Form



$
\begin{equation}
\begin{aligned}

y - y_1 =& m (x - x_1)
\\
\\
y - 3 =& \frac{2}{3} (x - 2)
\\
\\
y - 3 =& \frac{2x - 4}{3}
\\
\\
y - 3 + 2 =& \frac{2x - 4}{3} + 3
\\
\\
y =& \frac{2x - 4 + 9}{3}
\\
\\
y =& \frac{2x + 5}{3}
\qquad \qquad \text{Equation of the tangent line at $(2,3)$}

\end{aligned}
\end{equation}
$