Certainly! Let's begin by analyzing the function $f(x) = \sqrt{\frac{4x^2 - 16x + 20}{x^2 - 9}}$.

Step 1: Rewrite the function for differentiation

We can rewrite $f(x)$ in terms of powers for simplicity: $f(x) = \left(\frac{4x^2 - 16x + 20}{x^2 - 9}\right)^{1/2}.$

Using the chain rule, the derivative of $f(x)$ will be: $f'(x) = \frac{1}{2} \left(\frac{4x^2 - 16x + 20}{x^2 - 9}\right)^{-1/2} \cdot \frac{d}{dx} \left(\frac{4x^2 - 16x + 20}{x^2 - 9}\right).$

Step 2: Differentiate the quotient inside

We apply the quotient rule to differentiate $\frac{4x^2 - 16x + 20}{x^2 - 9}$. Let:

• $u(x) = 4x^2 - 16x + 20$,
• $v(x) = x^2 - 9$.

Using the quotient rule $\frac{u'v - vu'}{v^2}$, compute:

1. $u'(x) = \frac{d}{dx}(4x^2 - 16x + 20) = 8x - 16$,
2. $v'(x) = \frac{d}{dx}(x^2 - 9) = 2x$.

Substitute into the rule: $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{(8x - 16)(x^2 - 9) - (4x^2 - 16x + 20)(2x)}{(x^2 - 9)^2}.$

Step 3: Simplify the numerator

Expand and combine terms for the numerator: $(8x - 16)(x^2 - 9) = 8x^3 - 72x - 16x^2 + 144,$ $(4x^2 - 16x + 20)(2x) = 8x^3 - 32x^2 + 40x.$

The numerator becomes: $(8x^3 - 72x - 16x^2 + 144) - (8x^3 - 32x^2 + 40x) = -16x^2 + 144 - 32x^2 + 40x - 72x.$

Combine like terms: $-48x^2 - 32x + 144.$

Thus, the derivative of the quotient is: $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{-48x^2 - 32x + 144}{(x^2 - 9)^2}.$

Step 4: Substitute into $f'(x)$

Now substitute this result into the chain rule expression for $f'(x)$: $f'(x) = \frac{1}{2} \left(\frac{4x^2 - 16x + 20}{x^2 - 9}\right)^{-1/2} \cdot \frac{-48x^2 - 32x + 144}{(x^2 - 9)^2}.$

Simplify: $f'(x) = \frac{-48x^2 - 32x + 144}{2(x^2 - 9)^2 \sqrt{\frac{4x^2 - 16x + 20}{x^2 - 9}}}.$

Let me know if you'd like further simplification or explanation! 😊

Would you like me to:

1. Simplify this derivative further?
2. Explore critical points or zeros of $f'(x)$?
3. Analyze the domain or restrictions of $f(x)$?
4. Find higher-order derivatives?
5. Explain the quotient rule in greater detail?

Tip: Always check the domain of a function like this, as both the square root and denominator impose restrictions!