<s>[INST] 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' ={} & 4(-3)(6) \frac{1}{(6x+8)^4}
\\
y''={} & 4(-3)(-4)(6)^2 \frac{1}{(6x+8)^5}
\\
y'''={} & 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.
 [/INST] Notice, the series for product up to $n$th term $-3, -4, -5, -6, \ldots -(n+2)$ 
the series for power of $(6x+8)$ up to $n$th term is $4, 5, 6, \ldots (n+3)$ 
hence, the $n$th derivative can be generalized as follows $$y^n=4(-3)(-4)(-5)\ldots (-(n+2))(6)^n\frac{1}{(6x+8)^{n+3}}=2(-1)^n\left(1\cdot 2\cdot 3\cdot 4\ldots (n+2)\right)(6)^n\frac{1}{(6x+8)^{n+3}}$$ $$=2(-1)^n(n+2)!(6)^n\frac{1}{(6x+8)^{n+3}}$$ 
$$\bbox[5pt, border:2.5pt solid #FF0000]{\color{blue}{y^n=\frac{2(-1)^n(n+2)!6^n}{(6x+8)^{n+3}}}}$$
</s>