The function given is:

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

We need to find its derivative $f'(x)$. This is a quotient of two functions, so we will use the quotient rule for derivatives, which states:

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

Here,

• $u(x) = \sqrt{x^2 + 4}$
• $v(x) = \sqrt{x^2 - 4}$

Step 1: Derivatives of $u(x)$ and $v(x)$

• $u(x) = (x^2 + 4)^{1/2}$ $u'(x) = \frac{1}{2}(x^2 + 4)^{-1/2} \cdot 2x = \frac{x}{\sqrt{x^2 + 4}}$

• $v(x) = (x^2 - 4)^{1/2}$ $v'(x) = \frac{1}{2}(x^2 - 4)^{-1/2} \cdot 2x = \frac{x}{\sqrt{x^2 - 4}}$

Step 2: Apply the Quotient Rule

Now, using the quotient rule: $f'(x) = \frac{\frac{x}{\sqrt{x^2 + 4}} \cdot \sqrt{x^2 - 4} - \sqrt{x^2 + 4} \cdot \frac{x}{\sqrt{x^2 - 4}}}{(\sqrt{x^2 - 4})^2}$

Step 3: Simplify

• The denominator simplifies to: $(\sqrt{x^2 - 4})^2 = x^2 - 4$

• The numerator becomes: $\frac{x \sqrt{x^2 - 4}}{\sqrt{x^2 + 4}} - \frac{x \sqrt{x^2 + 4}}{\sqrt{x^2 - 4}}$

Combining the terms under a common denominator: $\text{Numerator} = \frac{x \left( (\sqrt{x^2 - 4})^2 - (\sqrt{x^2 + 4})^2 \right)}{\sqrt{x^2 + 4} \sqrt{x^2 - 4}} = \frac{x \left( (x^2 - 4) - (x^2 + 4) \right)}{\sqrt{x^2 + 4} \sqrt{x^2 - 4}}$

Simplifying further: $\text{Numerator} = \frac{x(-8)}{\sqrt{x^2 + 4} \sqrt{x^2 - 4}} = \frac{-8x}{\sqrt{x^2 + 4} \sqrt{x^2 - 4}}$

Final Answer:

The derivative of the function is: $f'(x) = \frac{-8x}{(x^2 - 4)\sqrt{x^2 + 4}}$

Let me know if you'd like any further clarification or detailed steps.

Related questions:

1. How do you apply the chain rule to derivatives of composite functions like $\sqrt{x^2 + 4}$?
2. What are some practical uses of the quotient rule in calculus?
3. How do you simplify expressions with square roots in both the numerator and denominator?
4. Can the derivative be used to analyze critical points of this function?
5. How does this derivative behave as $x$ approaches 2 or -2 (where the denominator becomes 0)?

Tip: Always check for points where the function or its derivative might be undefined, like where the denominator is zero.