Given that $y = (x+6)^7(x-9)^8$, can we find $dy/dx$?

Given that $y = (x+6)^7(x-9)^8$, can we find $dy/dx$?

Question

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.

Solution

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

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