## Problem Statement

Find the $n$-th order derivative.

$y=\frac{7 x+1}{17(4 x+3)}$
## Solution

$y=\frac{7 x+1}{17(4 x+3)}$

$y^{\prime}=\left(\frac{7 x+1}{17(4 x+3)}\right)^{\prime}=\frac{7 \cdot(4 x+3)-(7 x+1) \cdot 4}{17(4 x+3)^{2}}=$

$=\frac{28 x+21-28 x-4}{17(4 x+3)^{2}}=\frac{17}{17(4 x+3)^{2}}=\frac{1}{(4 x+3)^{2}}$

$y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=\left(\frac{1}{(4 x+3)^{2}}\right)^{\prime}=\frac{-2}{(4 x+3)^{3}} \cdot 4=-\frac{2 \cdot 4}{(4 x+3)^{3}}$

$y^{\prime \prime \prime}=\left(y^{\prime \prime}\right)^{\prime}=\left(-\frac{2 \cdot 4}{(4 x+3)^{3}}\right)^{\prime}=-\frac{2 \cdot(-3) \cdot 4}{(4 x+3)^{4}} \cdot 4=\frac{2 \cdot 3 \cdot 4^{2}}{(4 x+3)^{4}}$

...

Obviously,

$y^{(n)}=\frac{(-1)^{n-1} \cdot n!\cdot 4^{n-1}}{(4 x+3)^{n+1}}$