The problem asks to find the first and second derivatives of the function:

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

Let's go through each step.

Step 1: Find $f'(x)$

To differentiate $f(x) = \frac{x^2 - 1}{x^2 + 1}$, we use the quotient rule, which states that if $f(x) = \frac{g(x)}{h(x)}$, then: $f'(x) = \frac{g'(x) h(x) - g(x) h'(x)}{[h(x)]^2}$

In this case:

• $g(x) = x^2 - 1$ and $h(x) = x^2 + 1$
• $g'(x) = 2x$ and $h'(x) = 2x$

Applying the quotient rule: $f'(x) = \frac{(2x)(x^2 + 1) - (x^2 - 1)(2x)}{(x^2 + 1)^2}$

Simplify the numerator: $= \frac{2x(x^2 + 1) - 2x(x^2 - 1)}{(x^2 + 1)^2}$

Expanding each term in the numerator: $= \frac{(2x^3 + 2x) - (2x^3 - 2x)}{(x^2 + 1)^2}$

Combine like terms: $= \frac{2x^3 + 2x - 2x^3 + 2x}{(x^2 + 1)^2} = \frac{4x}{(x^2 + 1)^2}$

So, the first derivative is: $f'(x) = \frac{4x}{(x^2 + 1)^2}$

Step 2: Find $f''(x)$

To find the second derivative $f''(x)$, we need to differentiate $f'(x) = \frac{4x}{(x^2 + 1)^2}$ again, using the quotient rule.

Let:

• $g(x) = 4x$
• $h(x) = (x^2 + 1)^2$

Then:

• $g'(x) = 4$
• $h'(x) = 2(x^2 + 1)(2x) = 4x(x^2 + 1)$

Applying the quotient rule again: $f''(x) = \frac{g'(x) h(x) - g(x) h'(x)}{[h(x)]^2}$ $= \frac{4(x^2 + 1)^2 - 4x \cdot 4x(x^2 + 1)}{(x^2 + 1)^4}$

Simplify this expression: $= \frac{4(x^2 + 1)^2 - 16x^2(x^2 + 1)}{(x^2 + 1)^4}$

Factor out $4(x^2 + 1)$ from the numerator: $= \frac{4(x^2 + 1)(x^2 + 1 - 4x^2)}{(x^2 + 1)^4}$

Simplify inside the parentheses: $= \frac{4(x^2 + 1)(1 - 3x^2)}{(x^2 + 1)^3}$

So, the second derivative is: $f''(x) = \frac{4(1 - 3x^2)}{(x^2 + 1)^3}$

Final Answer:

1. First derivative: $f'(x) = \frac{4x}{(x^2 + 1)^2}$
2. Second derivative: $f''(x) = \frac{4(1 - 3x^2)}{(x^2 + 1)^3}$

Would you like more details on any step?

Here are some related questions to explore:

1. How can the quotient rule be used in real-life applications?
2. What is the physical significance of the second derivative in motion?
3. How would you find higher-order derivatives for this function?
4. How does the behavior of the function change at critical points of $f'(x)$?
5. Can $f(x)$ have any points of inflection, and how would you determine them?

Tip: Remember, the quotient rule is particularly useful for rational functions but can often be avoided by simplifying expressions first, if possible.