## Task Condition

Find the $n$-th order derivative.

$y=\lg (2 x+7)$
## Solution

$y^{\prime}=(\lg (2 x+7))^{\prime}=\frac{2}{(2 x+7) \ln 10}=\frac{2}{\ln 10} \cdot(2 x+7)^{-1}$

$y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=\left(\frac{2}{\ln 10} \cdot(2 x+7)^{-1}\right)^{\prime}=-\frac{2^{2}}{\ln 10} \cdot(2 x+7)^{-2}$

$y^{\prime \prime \prime}=\left(y^{\prime \prime}\right)^{\prime}=\left(-\frac{2^{2}}{\ln 10} \cdot(2 x+7)^{-2}\right)^{\prime}=\frac{2^{3} \cdot 2}{\ln 10} \cdot(2 x+7)^{-3}$

$y^{(4)}=\left(y^{\prime \prime \prime}\right)^{\prime}=\left(\frac{2^{3} \cdot 2}{\ln 10} \cdot(2 x+7)^{-3}\right)^{\prime}=-\frac{2^{4} \cdot 2 \cdot 3}{\ln 10} \cdot(2 x+7)^{-4}$

...

It is obvious that

$y^{(n)}=(-1)^{n-1} \cdot \frac{2^{n} \cdot(n-1)!}{\ln 10} \cdot(2 x+7)^{-n}$

## Problem Kuznetsov Differentiation 18-25