To solve problem 52, we need to find the ninth derivative of \( x^8 \ln x \):

\[ \frac{d^9}{dx^9} (x^8 \ln x) \]

First, let's express the general approach for finding the \( n \)-th derivative of a product of functions using the Leibniz rule for differentiation. The \( n \)-th derivative of a product of two functions \( u(x) \) and \( v(x) \) is given by:

\[ \frac{d^n}{dx^n} [u(x) v(x)] = \sum_{k=0}^{n} \binom{n}{k} \frac{d^{n-k}}{dx^{n-k}} [u(x)] \cdot \frac{d^k}{dx^k} [v(x)] \]

Here, \( u(x) = x^8 \) and \( v(x) = \ln x \). 

To apply this, we need to compute the derivatives of \( u(x) \) and \( v(x) \) up to the ninth derivative.

### Step-by-Step Solution

1. **Derivatives of \( u(x) = x^8 \)**:

   - \( u(x) = x^8 \)
   - \( u'(x) = 8x^7 \)
   - \( u''(x) = 56x^6 \)
   - \( u^{(3)}(x) = 336x^5 \)
   - \( u^{(4)}(x) = 1680x^4 \)
   - \( u^{(5)}(x) = 6720x^3 \)
   - \( u^{(6)}(x) = 20160x^2 \)
   - \( u^{(7)}(x) = 40320x \)
   - \( u^{(8)}(x) = 40320 \)
   - \( u^{(9)}(x) = 0 \) (since the 9th derivative of \( x^8 \) is zero)

2. **Derivatives of \( v(x) = \ln x \)**:

   - \( v(x) = \ln x \)
   - \( v'(x) = \frac{1}{x} \)
   - \( v''(x) = -\frac{1}{x^2} \)
   - \( v^{(3)}(x) = \frac{2}{x^3} \)
   - \( v^{(4)}(x) = -\frac{6}{x^4} \)
   - \( v^{(5)}(x) = \frac{24}{x^5} \)
   - \( v^{(6)}(x) = -\frac{120}{x^6} \)
   - \( v^{(7)}(x) = \frac{720}{x^7} \)
   - \( v^{(8)}(x) = -\frac{5040}{x^8} \)
   - \( v^{(9)}(x) = \frac{40320}{x^9} \)

3. **Applying the Leibniz Rule**:

   Using the Leibniz rule for \( n = 9 \):

   \[
   \frac{d^9}{dx^9} [x^8 \ln x] = \sum_{k=0}^{9} \binom{9}{k} \frac{d^{9-k}}{dx^{9-k}} [x^8] \cdot \frac{d^k}{dx^k} [\ln x]
   \]

   Since \( u^{(9)}(x) = 0 \) for \( k = 0 \) to \( 8 \), the only non-zero term occurs when \( k = 9 \):

   \[
   \frac{d^9}{dx^9} [x^8 \ln x] = \binom{9}{9} \frac{d^{9-9}}{dx^{9-9}} [x^8] \cdot \frac{d^9}{dx^9} [\ln x]
   \]

   Simplifying this:

   \[
   = 1 \cdot x^8 \cdot \frac{40320}{x^9}
   \]

   \[
   = 40320 \cdot \frac{x^8}{x^9}
   \]

   \[
   = 40320 \cdot \frac{1}{x}
   \]

   \[
   = \frac{40320}{x}
   \]

Therefore, the ninth derivative of \( x^8 \ln x \) is:

\[
\frac{d^9}{dx^9} (x^8 \ln x) = \frac{40320}{x}
\]

Would you like further details or have any questions?

Here are 5 more questions you might want to explore:

1. How do you find the integral of \( e^{x^2} \)?
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**Tip:** When calculating higher-order derivatives, identifying patterns in lower-order derivatives can simplify the process and help in verifying results.

Learn how to calculate the ninth derivative of x^8 ln x using the Leibniz rule for differentiation. This problem involves differentiating a product of functions and requires understanding of higher-order derivatives and logarithmic differentiation techniques. Suitable for advanced undergraduate students and anyone interested in calculus and mathematical analysis.

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