<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] $y=\frac{4}{(6x+8)^3} y'$ 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.
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