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We are given the function \( f(x) = \frac{(6x^2 - 5)(x^2 - 6)}{x^2 + 6} \), and we need to evaluate its derivative \( f'(x) \) at two points: \( f'(2) \) and \( f'(4) \). This involves using both the product rule and the quotient rule.

Let's walk through the steps to solve it:

### Step 1: Differentiate \( f(x) \) using the quotient rule
The quotient rule is given by:

\[
\left( \frac{u(x)}{v(x)} \right)' = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}
\]

Here, 
- \( u(x) = (6x^2 - 5)(x^2 - 6) \) (we will need to apply the product rule to differentiate this),
- \( v(x) = x^2 + 6 \).

### Step 2: Differentiate \( u(x) \) using the product rule
The product rule is given by:

\[
(uv)' = u'v + uv'
\]

Let:
- \( u_1(x) = 6x^2 - 5 \),
- \( u_2(x) = x^2 - 6 \).

Now, we find the derivatives:
\[
u_1'(x) = 12x, \quad u_2'(x) = 2x
\]

Using the product rule for \( u(x) = u_1(x) u_2(x) \):

\[
u'(x) = u_1'(x) u_2(x) + u_1(x) u_2'(x)
\]
\[
u'(x) = (12x)(x^2 - 6) + (6x^2 - 5)(2x)
\]
\[
u'(x) = 12x(x^2 - 6) + 2x(6x^2 - 5)
\]

Simplify:
\[
u'(x) = 12x^3 - 72x + 12x^3 - 10x
\]
\[
u'(x) = 24x^3 - 82x
\]

### Step 3: Differentiate \( v(x) \)
For \( v(x) = x^2 + 6 \), we have:
\[
v'(x) = 2x
\]

### Step 4: Apply the quotient rule
Now, applying the quotient rule to \( f(x) = \frac{u(x)}{v(x)} \):

\[
f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}
\]
Substitute the values of \( u'(x) \), \( v(x) \), and \( v'(x) \):

\[
f'(x) = \frac{(24x^3 - 82x)(x^2 + 6) - (6x^2 - 5)(x^2 - 6)(2x)}{(x^2 + 6)^2}
\]

### Step 5: Evaluate at \( x = 2 \) and \( x = 4 \)

Now, we can plug in \( x = 2 \) and \( x = 4 \) to get the specific values of the derivative at these points.

Let me compute these values.The derivative \( f'(x) \) at the given points evaluates as:

- \( f'(2) = \frac{108}{25} = 4.32 \)
- \( f'(4) = \frac{4824}{121} \approx 39.87 \)

Would you like further details or explanations?

Here are 5 related questions you can explore:
1. How do you apply the product rule when differentiating more complex products?
2. Why does the quotient rule involve squaring the denominator?
3. What happens to the quotient rule when \( v(x) \) is a constant?
4. How does simplification affect the accuracy of derivatives in applied problems?
5. How do critical points relate to where the derivative is zero?

**Tip:** Always check the domain of the function when using the quotient rule to ensure the denominator doesn't equal zero.

Evaluate f'(x) at the following points: (A) f'(2) = ?, (B) f'(4) = ?, where f(x) = [(6x^2 - 5)(x^2 - 6)] / (x^2 + 6)

Learn how to differentiate the function f(x) = [(6x^2 - 5)(x^2 - 6)] / (x^2 + 6) using both product and quotient rules. This detailed solution shows step-by-step differentiation at the points x = 2 and x = 4, providing critical insight into advanced calculus concepts.

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