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Let's solve problems 16 and 21:

### Problem 16
Given the function: 
\[
y = \frac{2x + 5}{2 - x}
\]
We are asked to find the derivative and evaluate the slope of the tangent line at \( x = -2 \).

**Step 1: Find the derivative \( \frac{dy}{dx} \).**

This is a quotient, so we use the **quotient rule**:
\[
\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}
\]
For our function:
- \( u = 2x + 5 \) and \( v = 2 - x \)
- \( \frac{du}{dx} = 2 \) and \( \frac{dv}{dx} = -1 \)

Applying the quotient rule:
\[
\frac{dy}{dx} = \frac{(2 - x)(2) - (2x + 5)(-1)}{(2 - x)^2}
\]
Simplify:
\[
\frac{dy}{dx} = \frac{4 - 2x + 2x + 5}{(2 - x)^2} = \frac{9}{(2 - x)^2}
\]

**Step 2: Evaluate the derivative at \( x = -2 \).**
\[
\frac{dy}{dx} \bigg|_{x = -2} = \frac{9}{(2 - (-2))^2} = \frac{9}{4^2} = \frac{9}{16}
\]

The slope of the tangent line at \( x = -2 \) is \( \frac{9}{16} \).

### Problem 21
Given the function:
\[
r = \frac{5}{\sqrt{9 - \theta}}
\]
We are asked to find \( \frac{dr}{d\theta} \) at \( \theta = 0 \).

**Step 1: Find the derivative \( \frac{dr}{d\theta} \).**

Rewrite the function:
\[
r = 5 \cdot (9 - \theta)^{-\frac{1}{2}}
\]

Use the **power rule** and the chain rule:
\[
\frac{dr}{d\theta} = 5 \cdot \left( -\frac{1}{2} \right) (9 - \theta)^{-\frac{3}{2}} \cdot (-1)
\]
Simplify:
\[
\frac{dr}{d\theta} = \frac{5}{2} (9 - \theta)^{-\frac{3}{2}}
\]

**Step 2: Evaluate the derivative at \( \theta = 0 \).**
\[
\frac{dr}{d\theta} \bigg|_{\theta = 0} = \frac{5}{2} (9 - 0)^{-\frac{3}{2}} = \frac{5}{2 \cdot 27} = \frac{5}{54}
\]

The value of \( \frac{dr}{d\theta} \) at \( \theta = 0 \) is \( \frac{5}{54} \).

Do you want more details or have any questions?

Here are some related questions to expand your understanding:

1. How would the slope change if we evaluated problem 16 at a different point?
2. What are the key steps for applying the quotient rule?
3. How does the chain rule help in problem 21?
4. What would happen to the derivative if the functions in problems 16 and 21 were squared?
5. How do we determine if a tangent line is horizontal?

**Tip:** Remember that the quotient rule requires careful application of the numerator and denominator derivatives—keeping track of signs is crucial!

Solve 21 and 16 from the provided math problem set.

Explore solutions to two calculus problems involving differentiation and tangent slopes. Problem 16 covers the differentiation of a quotient function, finding the tangent line's slope using the quotient rule. Problem 21 involves differentiating a radical function using the chain rule. Suitable for advanced high school calculus students.

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