Question

b. \\(\frac{\frac{2 - x - h}{1 + x + h} - \frac{2 - x}{1 + x}}{h}\\)

Explanation:

Step1: Simplify the numerator

First, we simplify the numerator \(\frac{2 - x - h}{1 + x + h}-\frac{2 - x}{1 + x}\). Find a common denominator \((1 + x + h)(1 + x)\):
\[

$$\begin{align*} &\frac{(2 - x - h)(1 + x)-(2 - x)(1 + x + h)}{(1 + x + h)(1 + x)}\\ =&\frac{(2(1 + x)-x(1 + x)-h(1 + x))-(2(1 + x + h)-x(1 + x + h))}{(1 + x + h)(1 + x)}\\ =&\frac{2 + 2x - x - x^{2}-h - hx - 2 - 2x - 2h + x + x^{2}+hx}{(1 + x + h)(1 + x)}\\ =&\frac{(2 - 2)+(2x - x - 2x + x)+(-x^{2}+x^{2})+(-h - 2h)+(-hx + hx)}{(1 + x + h)(1 + x)}\\ =&\frac{-3h}{(1 + x + h)(1 + x)} \end{align*}$$

\]

Step2: Divide by \(h\)

Now, we have the original expression as \(\frac{\frac{-3h}{(1 + x + h)(1 + x)}}{h}\). Dividing by \(h\) is the same as multiplying by \(\frac{1}{h}\):
\[
\frac{-3h}{(1 + x + h)(1 + x)}\times\frac{1}{h}=\frac{-3}{(1 + x + h)(1 + x)}
\]

Answer:

\(\frac{-3}{(1 + x)(1 + x + h)}\)