The provided image represents a derivative $f'(x)$ with a complex fraction involving functions of $x$, their derivatives, and powers. Here's how to analyze and simplify it step by step:

Denominator:

The denominator is $\left( x^2 - 6 \right)^6)^2 = \left( x^2 - 6 \right)^{12}$.

Numerator:

The numerator involves two terms:

1. $(x^2 - 6)^6 \cdot \frac{d}{dx}\left( 5 - x \right)$
2. $- (5 - x) \cdot \frac{d}{dx} \left( (x^2 - 6)^6 \right)$

Simplifying each term:

1. First term:

   • $\frac{d}{dx}(5 - x) = -1$, so the first term becomes: $(x^2 - 6)^6 \cdot (-1) = -(x^2 - 6)^6.$

2. Second term:

   • Using the chain rule for $\frac{d}{dx}\left( (x^2 - 6)^6 \right)$: $\frac{d}{dx}\left( (x^2 - 6)^6 \right) = 6 \cdot (x^2 - 6)^5 \cdot \frac{d}{dx}(x^2 - 6) = 6 \cdot (x^2 - 6)^5 \cdot 2x = 12x \cdot (x^2 - 6)^5.$
   • The second term becomes: $-(5 - x) \cdot 12x \cdot (x^2 - 6)^5.$

Combining the terms:

The numerator is: $-(x^2 - 6)^6 - (5 - x) \cdot 12x \cdot (x^2 - 6)^5.$

Full expression for $f'(x)$:

Now the derivative $f'(x)$ is: $f'(x) = \frac{-(x^2 - 6)^6 - (5 - x) \cdot 12x \cdot (x^2 - 6)^5}{\left( x^2 - 6 \right)^{12}}.$

Simplifying further:

Factor out $(x^2 - 6)^5$ from the numerator: $f'(x) = \frac{(x^2 - 6)^5 \left[ -(x^2 - 6) - 12x(5 - x) \right]}{\left( x^2 - 6 \right)^{12}}.$

Cancel $(x^2 - 6)^5$ from numerator and denominator: $f'(x) = \frac{-(x^2 - 6) - 12x(5 - x)}{\left( x^2 - 6 \right)^7}.$

Distribute $-12x$ in the numerator: $f'(x) = \frac{-(x^2 - 6) - 60x + 12x^2}{\left( x^2 - 6 \right)^7}.$

Combine like terms in the numerator: $f'(x) = \frac{-x^2 + 12x^2 - 60x + 6}{\left( x^2 - 6 \right)^7}.$

Simplify: $f'(x) = \frac{11x^2 - 60x + 6}{\left( x^2 - 6 \right)^7}.$

This is the simplified expression for $f'(x)$.

Let me know if you'd like more details on any step!

Related Questions:

1. How do you apply the chain rule for complex functions?
2. What are the best practices for simplifying derivatives involving powers?
3. How would the solution change if a constant factor were added to the function?
4. Why is factoring useful when simplifying derivatives?
5. Can this derivative be evaluated at specific points (e.g., $x = 0$ or $x = 5$)?

Tip:

Always simplify powers and common factors early to reduce complexity in large derivative expressions