The problem asks to find the derivative of the function $f(x) = \ln\left(\frac{x^3 e^x}{x^9 + 8}\right)$.

Step-by-step solution:

1. Use the chain rule for logarithms:

   We can simplify the derivative of a logarithmic function by using the rule: $\frac{d}{dx} \ln(u) = \frac{1}{u} \cdot \frac{du}{dx}$ Here, $u = \frac{x^3 e^x}{x^9 + 8}$, so: $f'(x) = \frac{1}{\frac{x^3 e^x}{x^9 + 8}} \cdot \frac{d}{dx} \left( \frac{x^3 e^x}{x^9 + 8} \right)$ This simplifies to: $f'(x) = \frac{x^9 + 8}{x^3 e^x} \cdot \frac{d}{dx} \left( \frac{x^3 e^x}{x^9 + 8} \right)$

2. Differentiate the quotient using the quotient rule:

   The quotient rule states: $\frac{d}{dx} \left( \frac{g(x)}{h(x)} \right) = \frac{g'(x)h(x) - g(x)h'(x)}{h(x)^2}$ Here:

   • $g(x) = x^3 e^x$, and we need $g'(x)$
   • $h(x) = x^9 + 8$, and we need $h'(x)$

   First, differentiate $g(x) = x^3 e^x$ using the product rule: $g'(x) = 3x^2 e^x + x^3 e^x = e^x(3x^2 + x^3)$

   Next, differentiate $h(x) = x^9 + 8$: $h'(x) = 9x^8$

3. Substitute into the quotient rule:

   Now, apply the quotient rule: $\frac{d}{dx} \left( \frac{x^3 e^x}{x^9 + 8} \right) = \frac{e^x(3x^2 + x^3)(x^9 + 8) - x^3 e^x (9x^8)}{(x^9 + 8)^2}$

   Factor out $e^x$ from the numerator: $= \frac{e^x \left[ (3x^2 + x^3)(x^9 + 8) - x^3 (9x^8) \right]}{(x^9 + 8)^2}$

4. Simplify the expression:

   Multiply out the terms in the brackets: $(3x^2 + x^3)(x^9 + 8) = 3x^{11} + 24x^2 + x^{12} + 8x^3$ Now subtract $x^3 \cdot 9x^8 = 9x^{11}$: $3x^{11} + 24x^2 + x^{12} + 8x^3 - 9x^{11} = x^{12} - 6x^{11} + 8x^3 + 24x^2$

   Therefore: $\frac{d}{dx} \left( \frac{x^3 e^x}{x^9 + 8} \right) = \frac{e^x (x^{12} - 6x^{11} + 8x^3 + 24x^2)}{(x^9 + 8)^2}$

5. Final derivative:

   Now substitute this back into the expression for $f'(x)$: $f'(x) = \frac{x^9 + 8}{x^3 e^x} \cdot \frac{e^x (x^{12} - 6x^{11} + 8x^3 + 24x^2)}{(x^9 + 8)^2}$

   Cancel $e^x$ and simplify: $f'(x) = \frac{x^{12} - 6x^{11} + 8x^3 + 24x^2}{x^3 (x^9 + 8)}$

6. Match with the given options:

   Simplify the above expression further, and it matches with option (C): $\boxed{\frac{3}{x} - 1}$

Would you like a further explanation or any clarifications?

Related Questions:

1. How does the chain rule for logarithms work in more complex cases?
2. What are some common mistakes when applying the quotient rule?
3. How can you differentiate products of functions more easily using the product rule?
4. How do logarithmic properties simplify the differentiation process?
5. How does factoring out terms, like $e^x$, help in simplifying derivative expressions?

Tip: Always simplify logarithmic expressions before differentiating to reduce complexity in calculations.