Calculus and Its Applications, Chapter 1, 1.7, Section 1.7, Problem 50

Determine $\displaystyle \frac{dy}{du}, \frac{du}{dx}$ and $\displaystyle \frac{dy}{dx}$ if $\displaystyle y = (u+1)(u-1)$ and $u = x^3 + 1$.

We first find $\displaystyle \frac{dy}{du}$ and $\displaystyle \frac{du}{dx}$.



$
\begin{equation}
\begin{aligned}

\frac{dy}{du} =& (u + 1) \cdot \frac{d}{du} (u-1) + (u-1) \cdot \frac{d}{du} (u + 1) \qquad \text{ and } &&& \frac{du}{dx} =& \frac
{d}{dx} (x^3) + \frac{d}{dx} (1)
\\
\\
=& (u+1)(1) + (u-1)(1) &&& =& 3x^2
\\
\\
=& u + 1 + u - 1
\\
\\
=& 2u

\end{aligned}
\end{equation}
$


Then,


$
\begin{equation}
\begin{aligned}

\frac{dy}{dx} =& \frac{dy}{du} \cdot \frac{du}{dx}
\\
\\
=& 2u \cdot 3x^2
\\
\\
=& 6ux^2
\\
\\
=& 6x^2 (x^3 + 1)
\qquad \text{Substitute $x^3 + 1$ for $u$}

\end{aligned}
\end{equation}
$