To find the derivative of the expression $f(x) = \frac{3 - x^2}{2 - x}$, we will use the quotient rule. The quotient rule is:

$\frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$

Here, let:

• $u(x) = 3 - x^2$
• $v(x) = 2 - x$

Step 1: Find $u'(x)$ and $v'(x)$

• $u'(x) = \frac{d}{dx} (3 - x^2) = -2x$
• $v'(x) = \frac{d}{dx} (2 - x) = -1$

Step 2: Apply the quotient rule

Using the quotient rule formula:

$f'(x) = \frac{(-2x)(2 - x) - (3 - x^2)(-1)}{(2 - x)^2}$

Step 3: Simplify the expression

1. Expand the numerator:

$(-2x)(2 - x) = -4x + 2x^2$ $(3 - x^2)(-1) = -3 + x^2$

So, the numerator becomes:

$(-4x + 2x^2) - (-3 + x^2) = -4x + 2x^2 + 3 - x^2 = x^2 - 4x + 3$

2. The denominator remains:

$(2 - x)^2 = (2 - x)(2 - x) = 4 - 4x + x^2$

Final answer:

The derivative is:

$f'(x) = \frac{x^2 - 4x + 3}{x^2 - 4x + 4}$

Let me know if you'd like more details or have any questions!

Here are 5 related questions:

1. How would you find the derivative of a product of two functions?
2. Can you simplify the final derivative further for specific values of $x$?
3. What happens to the derivative if $x = 2$ in this function?
4. How would the quotient rule change if both numerator and denominator had higher powers of $x$?
5. What is the second derivative of the given function?

Tip: Always simplify expressions before applying differentiation rules to avoid unnecessary complexity.