Hey! My friend sent me here 📸 SCANSOLVE_ZFB4WG

To find the derivative of $f(x) = \frac{x^2 - 1}{x^2 + 1}$, we will use the quotient rule. The quotient rule states that if $f(x) = \frac{g(x)}{h(x)}$, then $f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{(h(x))^2}$.

Step 1: Identify $g(x)$ and $h(x)$. Let $g(x) = x^2 - 1$. Let $h(x) = x^2 + 1$.

Step 2: Find the derivative of $g(x)$, denoted as $g'(x)$. $g'(x) = \frac{d}{dx}(x^2 - 1) = 2x$

Step 3: Find the derivative of $h(x)$, denoted as $h'(x)$. $h'(x) = \frac{d}{dx}(x^2 + 1) = 2x$

Step 4: Apply the quotient rule formula. $f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{(h(x))^2}$ Substitute the expressions for $g(x)$, $h(x)$, $g'(x)$, and $h'(x)$: $f'(x) = \frac{(2x)(x^2 + 1) - (x^2 - 1)(2x)}{(x^2 + 1)^2}$

Step 5: Expand and simplify the numerator. $Numerator = 2x(x^2 + 1) - (x^2 - 1)(2x)$ $Numerator = (2x^3 + 2x) - (2x^3 - 2x)$ $Numerator = 2x^3 + 2x - 2x^3 + 2x$ $Numerator = 4x$

Step 6: Write the final derivative. $f'(x) = \frac{4x}{(x^2 + 1)^2}$

The derivative of $f(x) = \frac{x^2 - 1}{x^2 + 1}$ is $\boxed{\frac{4x}{(x^2 + 1)^2}}$.