To solve the problem, we need to apply the quotient rule to find $h'(x)$, and then evaluate it at $x = 2$.

Given the function: $h(x) = \frac{-6x^2}{4x^2 + 6x - 10}$

Quotient Rule

The quotient rule states that if we have a function $h(x) = \frac{f(x)}{g(x)}$, then the derivative $h'(x)$ is given by: $h'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2}$ where:

• $f(x) = -6x^2$
• $g(x) = 4x^2 + 6x - 10$

Step 1: Compute the derivatives $f'(x)$ and $g'(x)$

$f'(x) = \frac{d}{dx}(-6x^2) = -12x$ $g'(x) = \frac{d}{dx}(4x^2 + 6x - 10) = 8x + 6$

Step 2: Apply the Quotient Rule

$h'(x) = \frac{(-12x)(4x^2 + 6x - 10) - (-6x^2)(8x + 6)}{(4x^2 + 6x - 10)^2}$ Simplifying the numerator: $= \frac{-48x^3 - 72x^2 + 120x + 48x^3 + 36x^2}{(4x^2 + 6x - 10)^2}$ $= \frac{-36x^2 + 120x}{(4x^2 + 6x - 10)^2}$

Step 3: Evaluate at $x = 2$

Substitute $x = 2$ into the simplified expression: $h'(2) = \frac{-36(2)^2 + 120(2)}{(4(2)^2 + 6(2) - 10)^2}$

Simplify: $h'(2) = \frac{-36(4) + 240}{(16 + 12 - 10)^2}$ $h'(2) = \frac{-144 + 240}{(18)^2}$ $h'(2) = \frac{96}{324}$ $h'(2) = \frac{8}{27}$

So, the exact value of $h'(2)$ is $\frac{8}{27}$.

Would you like any further details on this problem, or do you have any other questions?

Here are some related questions to expand your understanding:

1. How does the quotient rule differ from the product rule in calculus?
2. What is the significance of the numerator in the quotient rule?
3. Can the quotient rule be applied to functions with trigonometric expressions?
4. How would the quotient rule apply if the denominator $g(x)$ was a linear function?
5. What is the physical interpretation of $h'(x)$ in real-world problems?

Tip: Always simplify the expressions for the derivative as much as possible before substituting the value of $x$ to avoid unnecessary complex calculations.