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.