Given that \(y = (x+6)^7(x-9)^8\), find \(\dfrac{dy}{dx}\) and the values of \(x\) for which \(\dfrac{dy}{dx}\) is zero.

Let

\(u = (x+6)^7\) and

\(v = (x-9)^8\). By the

chain rule
\[\begin{equation*}
u' = 7(x+6)^6 \frac{d}{dx}(x+6) = 7(x+6)^6
\end{equation*}\]
and

\[\begin{equation*}
v' = 8(x-9)^7 \frac{d}{dx}(x-9) = 8(x-9)^7
\end{equation*}\]
By the

product rule
\[\begin{align*}
\frac{dy}{dx} = uv' + u'v &= 8(x+6)^7(x-9)^7 + 7(x+6)^6(x-9)^8 \\
&= (x+6)^6(x-9)^7 \left( 8(x+6) + 7(x-9) \right) \\
&= (x+6)^6(x-9)^7(15x - 15) \\
&= 15(x+6)^6(x-9)^7(x-1).
\end{align*}\]
Consequently,

\[\begin{equation*}
\frac{dy}{dx} = 0
\end{equation*}\]
if and only if

\[\begin{equation*}
x = -6 \quad\text{or}\quad x = 9 \quad\text{or}\quad x = 1.
\end{equation*}\]