The problem is:

$h(x) = \frac{2g(x)}{5 + f(x)}$

We are given the following values:

• $f(2) = -1$
• $f'(2) = 2$
• $g(2) = -2$
• $g'(2) = 4$

We need to find $h'(2)$, which involves using the quotient rule for differentiation.

Step 1: Apply the Quotient Rule

The quotient rule for a function $h(x) = \frac{u(x)}{v(x)}$ is given by: $h'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{v(x)^2}$ Here,

• $u(x) = 2g(x)$
• $v(x) = 5 + f(x)$

Step 2: Find the Derivatives

• $u'(x) = 2g'(x)$ (since $u(x) = 2g(x)$, we differentiate the constant factor)
• $v'(x) = f'(x)$ (since $v(x) = 5 + f(x)$, and the constant 5 disappears when differentiating)

Step 3: Substitute into the Quotient Rule

$h'(x) = \frac{(2g'(x))(5 + f(x)) - (2g(x))(f'(x))}{(5 + f(x))^2}$

Step 4: Substitute Given Values at $x = 2$

We are given:

• $f(2) = -1$
• $f'(2) = 2$
• $g(2) = -2$
• $g'(2) = 4$

Substitute these into the expression for $h'(2)$: $h'(2) = \frac{(2 \times 4)(5 + (-1)) - (2 \times (-2))(2)}{(5 + (-1))^2}$

Step 5: Simplify the Expression

First, simplify the terms inside the parentheses:

• $5 + (-1) = 4$

Now substitute into the equation: $h'(2) = \frac{(2 \times 4)(4) - (2 \times (-2))(2)}{(4)^2}$ $h'(2) = \frac{(8)(4) - (-4)(2)}{16}$ $h'(2) = \frac{32 + 8}{16}$ $h'(2) = \frac{40}{16} = 2.5$

Thus, the value of $h'(2)$ is 2.5.

Would you like further details or clarification on any step?

Here are 5 related questions you can explore:

1. How does the quotient rule apply to more complex functions?
2. What is the product rule, and how does it differ from the quotient rule?
3. How can the chain rule be used alongside the quotient rule?
4. How do second derivatives apply in a quotient-rule scenario?
5. What are real-world applications of the quotient rule in calculus?

Tip: Always simplify terms inside parentheses or fractions first when working with derivatives to reduce complexity.