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.
 A: $$\begin{align}
y' ={} & 4(-3)(6) \frac{1}{(6x+8)^4}=-\frac{3!}{2}6^1\frac{4}{(6x+8)^{3+1}}
\\
y''={} & 4(-3)(-4)(6)^2 \frac{1}{(6x+8)^5}=\frac{4!}{2}6^2\frac{4}{(6x+8)^{3+2}}
\\
y'''={} & 4(-3)(-4)(-5)(6)^3\frac{1}{(6x+8)^6}=-\frac{5!}{2}6^3\frac{4}{(6x+8)^{3+3}}
\end{align}$$
so, the formula is
$$y^{n}=(-1)^n\frac{(2+n)!}{2}6^n\frac{4}{(6x+8)^{3+n}}=(-1)^n(2+n)!6^n\frac{2}{(6x+8)^{3+n}}$$
A: 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 $$y^n=4(-3)(-4)(-5)\ldots (-(n+2))(6)^n\frac{1}{(6x+8)^{n+3}}=2(-1)^n\left(1\cdot 2\cdot 3\cdot 4\ldots (n+2)\right)(6)^n\frac{1}{(6x+8)^{n+3}}$$ $$=2(-1)^n(n+2)!(6)^n\frac{1}{(6x+8)^{n+3}}$$ 
$$\bbox[5pt, border:2.5pt solid #FF0000]{\color{blue}{y^n=\frac{2(-1)^n(n+2)!6^n}{(6x+8)^{n+3}}}}$$
A: The general term for nth derivative would be $(-1)^n.4.6^n.(3)..(3+(n-1)\frac{1}{(6x+8)^{n+3}}$. 'n' indicates the nth derivative.
A: $y=\frac{4}{(6x+8)^3} y'$ 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.
