Given the function f, of, x, equals, start fraction, 1, minus, x, divided by, 2, x, squared, minus, 5, end fraction, comma*f*(*x*)=2*x*2−51−*x*​, find f, prime, of, x*f*′(*x*) in *simplified* form.

Answer

Attempt 1 out of 2

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

Given the function f(x) = (1 - x) / (2x^2 - 5), find f'(x) in simplified form.

Discover how to calculate the derivative of the function f(x) = (1 - x) / (2x^2 - 5) using the quotient rule. This solution involves key calculus concepts like differentiation and the application of the quotient rule. Ideal for students in Grades 11-12 or early university.

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