The function you provided is:

$f(x) = \frac{4 - 2x - x^2}{x^2 - 6}$

We need to find the derivative of this function $f(x)$, which is a quotient of two functions. To differentiate, we use the quotient rule:

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

Here,

• $u(x) = 4 - 2x - x^2$
• $v(x) = x^2 - 6$

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

• $u'(x) = \frac{d}{dx}(4 - 2x - x^2) = -2 - 2x$
• $v'(x) = \frac{d}{dx}(x^2 - 6) = 2x$

Step 2: Apply the quotient rule

$f'(x) = \frac{(x^2 - 6)(-2 - 2x) - (4 - 2x - x^2)(2x)}{(x^2 - 6)^2}$

Step 3: Simplify the expression

Let's expand both parts of the numerator:

1. Expanding $(x^2 - 6)(-2 - 2x)$: $= x^2(-2 - 2x) - 6(-2 - 2x) = -2x^2 - 2x^3 + 12 + 12x$ So, $(x^2 - 6)(-2 - 2x) = -2x^3 - 2x^2 + 12x + 12$.

2. Expanding $(4 - 2x - x^2)(2x)$: $= 4(2x) - 2x(2x) - x^2(2x) = 8x - 4x^2 - 2x^3$ So, $(4 - 2x - x^2)(2x) = 8x - 4x^2 - 2x^3$.

Step 4: Combine the results in the numerator

Now, subtract the two expanded terms: $(-2x^3 - 2x^2 + 12x + 12) - (8x - 4x^2 - 2x^3)$ This simplifies to: $-2x^3 - 2x^2 + 12x + 12 - 8x + 4x^2 + 2x^3$ $= (0)x^3 + 2x^2 + 4x + 12$ So, the numerator becomes $2x^2 + 4x + 12$.

Step 5: Final expression for $f'(x)$

The derivative is: $f'(x) = \frac{2x^2 + 4x + 12}{(x^2 - 6)^2}$

This is the derivative of the function.

Would you like a further breakdown of any steps?

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Here are 5 related questions:

1. How would you apply the product rule in differentiation?
2. Can the chain rule be used with this function?
3. What is the interpretation of the quotient rule in calculus?
4. How do second derivatives relate to concavity in functions?
5. How does implicit differentiation differ from the explicit method?

Tip: Always double-check algebraic expansions to avoid small errors in multi-term expressions.