To find $\frac{dy}{dx}$ using the quotient rule for $y = \frac{1 + 3x}{2 - x}$, follow these steps:

The quotient rule states:

$\frac{dy}{dx} = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}$

Here:

• $f(x) = 1 + 3x$
• $g(x) = 2 - x$

Step 1: Differentiate $f(x)$ and $g(x)$

$f'(x) = \frac{d}{dx}(1 + 3x) = 3$ $g'(x) = \frac{d}{dx}(2 - x) = -1$

Step 2: Apply the quotient rule

Substitute into the formula: $\frac{dy}{dx} = \frac{(f'(x)g(x)) - (f(x)g'(x))}{[g(x)]^2}$

Substitute $f(x)$, $f'(x)$, $g(x)$, and $g'(x)$: $\frac{dy}{dx} = \frac{(3)(2 - x) - (1 + 3x)(-1)}{(2 - x)^2}$

Step 3: Simplify the numerator

Expand the terms in the numerator: $\frac{dy}{dx} = \frac{6 - 3x - (1 + 3x)(-1)}{(2 - x)^2}$

Simplify further: $\frac{dy}{dx} = \frac{6 - 3x + 1 + 3x}{(2 - x)^2}$

Combine like terms: $\frac{dy}{dx} = \frac{7}{(2 - x)^2}$

Final Answer:

$\frac{dy}{dx} = \frac{7}{(2 - x)^2}$

Would you like further clarification or detailed explanation?

Here are 5 related questions to enhance your understanding:

1. How does the quotient rule differ from the product rule in calculus?
2. Can you derive the same result by rewriting the function as $y = (1+3x)(2-x)^{-1}$ and using the chain rule?
3. What happens to $\frac{dy}{dx}$ if $x = 2$, and why does it cause a problem?
4. How can you verify the result of the quotient rule by plotting both $y$ and $\frac{dy}{dx}$?
5. What are the real-world applications of the quotient rule?

Tip: Always double-check your derivatives for both the numerator and denominator separately before combining them using the quotient rule!