Calculus and Its Applications, Chapter 1, 1.7, Section 1.7, Problem 46

Determine $\displaystyle \frac{dy}{du}, \frac{du}{dx}$ and $\displaystyle \frac{dy}{dx}$ if $\displaystyle y = \frac{15}{u^3}$ and $u = 2x + 1$.

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


$
\begin{equation}
\begin{aligned}

\frac{dy}{du} =& \frac{\displaystyle u^3 \cdot \frac{d}{du} (15) - 15 \cdot \frac{d}{du} (u^3)}{(u^3)^2} \qquad \text{ and } &&& \frac{du}{dx} =& 2 \frac{d}{dx} (x) + \frac{d}{dx} 1
\\
\\
=& \frac{-15 (3u^2)}{u^6} &&& =& 2
\\
\\
=& \frac{-45}{u^4} &&

\end{aligned}
\end{equation}
$


Then,


$
\begin{equation}
\begin{aligned}

\frac{dy}{dx} =& \frac{dy}{du} \cdot \frac{du}{dx}
\\
\\
=& \frac{-45}{u^4} \cdot 2
\\
\\
=& \frac{-90}{(2x+1)^4}
\qquad \text{Substitute $2x+1$ for $u$}

\end{aligned}
\end{equation}
$