Single Variable Calculus, Chapter 7, 7.4-1, Section 7.4-1, Problem 56

Determine $\displaystyle \frac{d^9}{dx^9} \left( x^8 \ln x \right)$

$
\begin{equation}
\begin{aligned}
\frac{d}{dx} \left( x^8 \ln x \right) &= x^8 \frac{d}{dx} (\ln x) + (\ln x) \frac{d}{dx} (x^8)\\
\\
\frac{d}{dx} \left( x^8 \ln x \right) &= x^8 \cdot \frac{1}{x} + \ln x \cdot 8 x^7\\
\\
\frac{d}{dx} \left( x^8 \ln x \right) &= x^7 + 8x^7 \ln x
\end{aligned}
\end{equation}
$


$
\begin{equation}
\begin{aligned}
\frac{d^2}{dx^2} \left( x^8 \ln x \right) &= \frac{d}{dx} (x^7) + \left[ 8x^7 \frac{d}{dx}(\ln x)+ (\ln x) 8 \frac{d}{dx} (x^7) \right]\\
\\
\frac{d^2}{dx^2} \left( x^8 \ln x \right) &= 7x^6 + 8x^7 \cdot \frac{1}{x} + \ln x \cdot 56 x^6\\
\\
\frac{d^2}{dx^2} \left( x^8 \ln x \right) &= 7x^6 + 8x^6 + 56x^6 \ln x\\
\\
\frac{d^2}{dx^2} \left( x^8 \ln x \right) &= 15x^6 + 56x^6 \ln x
\end{aligned}
\end{equation}
$


$
\begin{equation}
\begin{aligned}
\frac{d^3}{dx^3} \left( x^8 \ln x \right) &= 15 \frac{d}{dx} (x^6) + \left[ 56x^6 \frac{d}{dx} (\ln x) + (\ln x) 56 \frac{d}{dx}(x^6) \right]\\
\\
\frac{d^3}{dx^3} \left( x^8 \ln x \right) &= 15(6x^5) + 56x^6 \cdot \frac{1}{x} + \ln x \cdot 56(6x^5)\\
\\
\frac{d^3}{dx^3} \left( x^8 \ln x \right) &= 90x^5 + 56x^5 + 336x^5 \ln x\\
\\
\frac{d^3}{dx^3} \left( x^8 \ln x \right) &= 146x^5 + 336x^5 \ln x
\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}
\frac{d^4}{dx^4} \left( x^8 \ln x \right) &= 146 \frac{d}{dx} (x^5) + \left[ 336x^5 \frac{d}{dx} (\ln x) + (\ln x) 336 \frac{d}{dx}(x^5) \right]\\
\\
\frac{d^4}{dx^4} \left( x^8 \ln x \right) &= 146 (5x^4) + 336x^5 \cdot \frac{1}{x} + \ln x \cdot 336(5x^4)\\
\\
\frac{d^4}{dx^4} \left( x^8 \ln x \right) &= 730x^4 + 336x^4 + 1680 x^4 \ln x\\
\\
\frac{d^4}{dx^4} \left( x^8 \ln x \right) &= 1066 x^4 + 1680x^4 \ln x
\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}
\frac{d^5}{dx^5} \left( x^8 \ln x \right) &= 1066 \frac{d}{dx} (x^5) + \left[ 1680x^4 \frac{d}{dx} (\ln x) + (\ln x) 1680 \frac{d}{dx}(x^4) \right]\\
\\
\frac{d^5}{dx^5} \left( x^8 \ln x \right) &= 1066 (4x^3) + 1680x^4 \cdot \frac{1}{x} + \ln x \cdot 1680(4x^3)\\
\\
\frac{d^5}{dx^5} \left( x^8 \ln x \right) &= 4264x^3 + 1680x^3 + 6720 x^3 \ln x\\
\\
\frac{d^5}{dx^5} \left( x^8 \ln x \right) &= 5944 x^3 + 6720x^3 \ln x
\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}
\frac{d^6}{dx^6} \left( x^8 \ln x \right) &= 5944 \frac{d}{dx} (x^3) + \left[ 6720x^3 \frac{d}{dx} (\ln x) + (\ln x) 6720 \frac{d}{dx}(x^3) \right]\\
\\
\frac{d^6}{dx^6} \left( x^8 \ln x \right) &= 5944 (3x^2) + 6720 x^3 \cdot \frac{1}{x} + \ln x \cdot 6720(3x^2)\\
\\
\frac{d^6}{dx^6} \left( x^8 \ln x \right) &= 17832 x^2 + 6720 x^2 + 20160 x^2 \ln x\\
\\
\frac{d^6}{dx^6} \left( x^8 \ln x \right) &= 24552 x^2 + 20160 x^2 \ln x
\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}
\frac{d^7}{dx^7} \left( x^8 \ln x \right) &= 24552 \frac{d}{dx} (x^2) + \left[ 20160 x^2 \frac{d}{dx} (\ln x) + (\ln x) 20160 \frac{d}{dx}(x^2) \right]\\
\\
\frac{d^7}{dx^7} \left( x^8 \ln x \right) &= 24552 (2x) + 20160 x^2 \cdot \frac{1}{x} + \ln x \cdot 20160(2x)\\
\\
\frac{d^7}{dx^7} \left( x^8 \ln x \right) &= 49104 x + 20160 x + 40320 x\ln x\\
\\
\frac{d^7}{dx^7} \left( x^8 \ln x \right) &= 69264 x + 40320 x \ln x
\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}
\frac{d^8}{dx^8} \left( x^8 \ln x \right) &= 69264 \frac{d}{dx} (x) + \left[ 40320 x \frac{d}{dx} (\ln x) + (\ln x) 40320 \frac{d}{dx}(x) \right]\\
\\
\frac{d^8}{dx^8} \left( x^8 \ln x \right) &= 69264 + 40320 x \cdot \frac{1}{x} + 40320 \ln x\\
\\
\frac{d^8}{dx^8} \left( x^8 \ln x \right) &= 69264 + 40320 + 40320 \ln x\\
\\
\frac{d^8}{dx^8} \left( x^8 \ln x \right) &= 109584 + 40320 \ln x
\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}
\frac{d^9}{dx^9} \left( x^8 \ln x \right) &= \frac{d}{dx} (109584) + 40320 \frac{d}{dx}(\ln x)\\
\\
\frac{d^9}{dx^9} \left( x^8 \ln x \right) &= 0 + 40320 \cdot \frac{1}{x}
\\
\frac{d^9}{dx^9} \left( x^8 \ln x \right) &= \frac{40320}{x}
\end{aligned}
\end{equation}
$