We are asked to find the derivative of the function $f(x) = \frac{1 - x}{2x^2 - 5}$.

To do this, we'll use the quotient rule for derivatives, which is given by:

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

where:

• $u(x) = 1 - x$
• $v(x) = 2x^2 - 5$

Step 1: Differentiate $u(x)$ and $v(x)$

• $u'(x) = \frac{d}{dx}(1 - x) = -1$
• $v'(x) = \frac{d}{dx}(2x^2 - 5) = 4x$

Step 2: Apply the quotient rule

Using the quotient rule formula:

$f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}$

Substitute $u(x)$, $v(x)$, $u'(x)$, and $v'(x)$ into the formula:

$f'(x) = \frac{-1(2x^2 - 5) - (1 - x)(4x)}{(2x^2 - 5)^2}$

Step 3: Simplify the expression

First, expand both the numerator terms:

$f'(x) = \frac{- (2x^2 - 5) - (1 - x)(4x)}{(2x^2 - 5)^2}$ $f'(x) = \frac{-2x^2 + 5 - (4x - 4x^2)}{(2x^2 - 5)^2}$ $f'(x) = \frac{-2x^2 + 5 - 4x + 4x^2}{(2x^2 - 5)^2}$

Combine like terms:

$f'(x) = \frac{2x^2 - 4x + 5}{(2x^2 - 5)^2}$

Thus, the derivative of $f(x)$ is:

[ f'(x) = \frac