9. Let $$ f, g: \mathbb{R} \rightarrow \mathbb{R} $$ and suppose that $$ f(2)=-3, g(2)=4, f^{\prime}(2)=-2 $$, and $$ g^{\prime}(2)=7 $$. Determine $$ h^{\prime}(2) $$ if $$ h $$ is defined as follows: $$ h(x)=\frac{x^{2}+f(x)}{f(x) g(x)-x-1} $$

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We begin with
$$
h(x)=\frac{x^{2}+f(x)}{f(x)g(x)-x-1}.
$$
Let 
$$
N(x)=x^2+f(x) \quad \text{and} \quad D(x)=f(x)g(x)-x-1.
$$
Then by the quotient rule,
$$
h'(x)=\frac{N'(x)D(x)-N(x)D'(x)}{[D(x)]^2}.
$$

**Step 1. Compute $$N(2)$$ and $$N'(2)$$:**

The derivative of $$N(x)$$ is
$$
N'(x)=2x+f'(x).
$$
At $$x=2$$:
$$
N(2)=2^2+f(2)=4+(-3)=1,
$$
$$
N'(2)=2(2)+f'(2)=4+(-2)=2.
$$

**Step 2. Compute $$D(2)$$ and $$D'(2)$$:**

The denominator is given by
$$
D(x)=f(x)g(x)-x-1.
$$
Differentiate using the product rule:
$$
D'(x)=f'(x)g(x)+f(x)g'(x)-1.
$$
At $$x=2$$:
$$
D(2)=f(2)g(2)-2-1=(-3)(4)-3=-12-3=-15,
$$
$$
D'(2)=f'(2)g(2)+f(2)g'(2)-1=(-2)(4)+(-3)(7)-1=-8-21-1=-30.
$$

**Step 3. Apply the quotient rule at $$x=2$$:**

Substitute the computed values into the quotient rule formula:
$$
h'(2)=\frac{N'(2)D(2)-N(2)D'(2)}{[D(2)]^2}=\frac{2(-15)-1(-30)}{(-15)^2}.
$$
Simplify the numerator:
$$
2(-15)-1(-30)=-30+30=0.
$$
And the denominator:
$$
(-15)^2=225.
$$
Thus, we have:
$$
h'(2)=\frac{0}{225}=0.
$$

The final answer is $$ h'(2)=0 $$.
$$ h'(2) = 0 $$

Let and suppose that , and . Determine if is
defined as follows:

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