Single Variable Calculus, Chapter 7, 7.2-2, Section 7.2-2, Problem 50

Find $\displaystyle \frac{d^9}{dx^9} (x^8 \ln x)$.


$\begin{equation} \begin{aligned} & \text{if } f(x) = x^8 \ln x, \text{ then by using Product Rule.. } \\ \\ & \frac{d}{dx} = x^8 \left( \frac{1}{x} \right) + 8x^7 \ln x \\ \\ & \frac{d}{dx} = x^7 + 8x^7 \ln x \end{aligned} \end{equation}$


Again, by using Product Rule..


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


Again, by using Product Rule..


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


From these pattern, we can see that if you differentiate the $x^8$ nine times, the term will go to 0 since at the 8th time, the term is already a constant and we know that the derivative of a constant is 0. Thus, we can disregard the first term. Also, the second term is similar to the first term except that it is multiplied by 8 and the power is one less.

Now, after taking the derivative 7 times we get..

$\displaystyle \frac{d^7}{dx^7} = 8! \left( \ln (x) + \left( \frac{1}{x} \right) (x) \right) + \text{ constant}$

Therefore, the ninth domain is


$\begin{equation} \begin{aligned} \frac{d^9}{dx^9} =& 8! \left( \frac{\displaystyle \frac{d}{dx} (x)}{x} \right) = 8! \left( \frac{1}{x} \right) \\ \\ \frac{d^9}{dx^9} =& \frac{8!}{x} \end{aligned} \end{equation}$