Calculus and Its Applications, Chapter 1, 1.7, Section 1.7, Problem 56

Determine $\displaystyle \frac{dy}{dt}$ if $\displaystyle y = \frac{1}{3u^5 - 7}$ and $u = 7t^2 + 1$.

We have $\displaystyle \frac{dy}{dt} = \frac{dy}{du} \cdot \frac{du}{dt}$
with


$
\begin{equation}
\begin{aligned}

\frac{dy}{du} =& \frac{\displaystyle (3u^5 - 7) \cdot \frac{d}{du} (1) - 1 \cdot \frac{d}{du} (3u^5 - 7) }{(3u^5 - 7)^2} \qquad \text{ and } &&& \frac{du}{dt} =& 7 \frac{d}{dt} (t^2) + \frac{d}{dt} 1
\\
=& \frac{-15u^4}{(3u^5 - 7)^2} &&& =& 14t

\end{aligned}
\end{equation}
$


Thus,


$
\begin{equation}
\begin{aligned}

\frac{dy}{dt} = \frac{-15u^4}{(3u^5 - 7)^2} \cdot 14t =& \frac{-210 tu^4}{(3u^5 - 7)^2}
\\
\\
=& \frac{-210 t (7t^2 + 1)^4}{[3(7t^2 + 1)^5 - 7 ]^2}
&& \text{Substitute $7t^2 + 1$ for $u$}

\end{aligned}
\end{equation}
$