Calculus and Its Applications, Chapter 1, 1.7, Section 1.7, Problem 58

Determine an equation for the tangent line to the graph of $y = (x^3 - 4x)^{10}$ at the point $(2,0)$.

We take the first derivative of the equation to find the slope,


$
\begin{equation}
\begin{aligned}

y' = m =& 10 (x^3 - 4x)^9 \cdot \frac{d}{dx} (x^3 - 4x)
\\
m =& 10 (x^3 - 4x)^9 (3x^2 - 4)
\\
m =& 10 [(2)^3 - 4(2)]^9 [3(2)^2 - 4] \qquad \text{Substitute } x=2
\\
m =& 10 (8-8)^9 (12-4)
\\
m =& 10(0) (8)
\\
m =& 0

\end{aligned}
\end{equation}
$


Using the Point Slope Form, the equation of the tangent line is


$
\begin{equation}
\begin{aligned}

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

\end{aligned}
\end{equation}
$