To show that $y = \log(1+x) - \frac{2x}{2+x}$ is an increasing function for $x > -1$, we need to prove that $\frac{dy}{dx} \geq 0$ throughout its domain.

Step 1: Find the derivative

Given: $y = \log(1+x) - \frac{2x}{2+x}$

Differentiating both terms:

• $\frac{d}{dx}[\log(1+x)] = \frac{1}{1+x}$
• $\frac{d}{dx}\left[\frac{2x}{2+x}\right] = \frac{2(2+x) - 2x(1)}{(2+x)^2} = \frac{4}{(2+x)^2}$

Therefore:

$$\frac{dy}{dx} = \frac{1}{1+x} - \frac{4}{(2+x)^2}$$

Step 2: Simplify the derivative

Taking common denominator:

$$\frac{dy}{dx} = \frac{(2+x)^2 - 4(1+x)}{(1+x)(2+x)^2}$$

Expanding the numerator:

$$(2+x)^2 = x^2 + 4x + 4$$ $$4(1+x) = 4 + 4x$$ $$\text{Numerator} = (x^2 + 4x + 4) - (4 + 4x) = x^2$$

Therefore:

$$\frac{dy}{dx} = \frac{x^2}{(1+x)(2+x)^2}$$

Step 3: Analyze the sign for $x > -1$

For all $x > -1$:

• $x^2 \geq 0$ (always non-negative)
• $(1+x) > 0$ (since $x > -1$)
• $(2+x)^2 > 0$ (always positive)

Since all factors in the denominator are positive and $x^2 \geq 0$, we have:

$$\frac{dy}{dx} \geq 0 \text{ for all } x > -1$$

Conclusion:

Since the derivative is non-negative throughout the domain $x > -1$, the function $y = \log(1+x) - \frac{2x}{2+x}$ is an increasing function throughout its domain.