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.


$\begin{equation} \begin{aligned} y - y_1 =& m (x - x_1) \\ \\ y - 0 =& 12 (x - 2) \\ \\ y =& 12 x - 24 \end{aligned} \end{equation}$