Solution

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