Single Variable Calculus, Chapter 7, 7.2-2, Section 7.2-2, Problem 46

Find an equation of the tangent line to the curve $y = \ln (x^3 - 7)$ at the point $(2, 0)$.


$
\begin{equation}
\begin{aligned}

\text{if } y =& \ln (x^3 - 7), \text{ then}
\\
\\
y' =& \frac{\displaystyle \frac{d}{dx} (x^3 - 7)}{x^3 - 7}
\\
\\
y' =& \frac{3x^2}{x^3 - 7}

\end{aligned}
\end{equation}
$


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

Thus, at point $(2, 0)$,


$
\begin{equation}
\begin{aligned}

y' =& \frac{3(2)^2}{2^3 - 7}
\\
\\
y' =& 12

\end{aligned}
\end{equation}
$


Therefore, the equation of the tangent line to the curve can be determined by using the point slope form.


$
\begin{equation}
\begin{aligned}

y - y_1 =& m (x - x_1)
\\
\\
y - 0 =& 12 (x - 2)
\\
\\
y =& 12 x - 24

\end{aligned}
\end{equation}
$