Question

1. if $y = \frac{1 - x}{2x + 1}$, then $\frac{dy}{dx} =$

Explanation:

Step1: Recall Quotient Rule

The quotient rule for differentiation states that if \( y = \frac{u}{v} \), then \( \frac{dy}{dx} = \frac{u'v - uv'}{v^2} \), where \( u \) and \( v \) are functions of \( x \), and \( u' \) and \( v' \) are their derivatives with respect to \( x \).

Let \( u = 1 - x \) and \( v = 2x + 1 \).

Step2: Find \( u' \) and \( v' \)

• For \( u = 1 - x \), the derivative \( u' \) with respect to \( x \) is \( u' = \frac{d}{dx}(1 - x) = -1 \).
• For \( v = 2x + 1 \), the derivative \( v' \) with respect to \( x \) is \( v' = \frac{d}{dx}(2x + 1) = 2 \).

Step3: Apply Quotient Rule

Substitute \( u \), \( v \), \( u' \), and \( v' \) into the quotient rule formula:
\[
\frac{dy}{dx} = \frac{u'v - uv'}{v^2} = \frac{(-1)(2x + 1) - (1 - x)(2)}{(2x + 1)^2}
\]

Step4: Simplify the Numerator

First, expand the numerator:
\[

$$\begin{align*} (-1)(2x + 1) - (1 - x)(2) &= -2x - 1 - 2 + 2x \\ &= (-2x + 2x) + (-1 - 2) \\ &= 0 - 3 \\ &= -3 \end{align*}$$

\]

Step5: Write the Final Derivative

Now, substitute the simplified numerator back into the quotient rule formula:
\[
\frac{dy}{dx} = \frac{-3}{(2x + 1)^2}
\]

Answer:

\( \frac{-3}{(2x + 1)^2} \)