Finding the nth derivative of $y=\frac4{(6x+8)^3}$ Find the $n$-th derivative of $y=\frac{4}{(6x+8)^3}$

\begin{align}
y' ={} &amp; 4(-3)(6) \frac{1}{(6x+8)^4}
\\
y''={} &amp; 4(-3)(-4)(6)^2 \frac{1}{(6x+8)^5}
\\
y'''={} &amp; 4(-3)(-4)(-5)(6)^3\frac{1}{(6x+8)^6}
\end{align}
I recognise the pattern but can't interpret that into a formula.
$$\begin{align}
y' ={} &amp; 4(-3)(6) \frac{1}{(6x+8)^4}=-\frac{3!}{2}6^1\frac{4}{(6x+8)^{3+1}}
\\
y''={} &amp; 4(-3)(-4)(6)^2 \frac{1}{(6x+8)^5}=\frac{4!}{2}6^2\frac{4}{(6x+8)^{3+2}}
\\
y'''={} &amp; 4(-3)(-4)(-5)(6)^3\frac{1}{(6x+8)^6}=-\frac{5!}{2}6^3\frac{4}{(6x+8)^{3+3}}
\end{align}$$
so, the formula is
$$y^{n}=(-1)^n\frac{(2+n)!}{2}6^n\frac{4}{(6x+8)^{3+n}}=(-1)^n(2+n)!6^n\frac{2}{(6x+8)^{3+n}}$$
$y=\frac{4}{(6x+8)^3} y'$ so $\frac{\rm d}{\rm dx}(\ln y)=\frac{(6x+8)^3}{4}$.
Now you can find $\ln y$ and hence $y$, and then take the derivatives.