Q.Show that $y = \log(1+x) - \frac{2x}{2+x}$, $x > -1$, is an increasing function of $x$ throughout its domain.

Concept: Increasing Function Test — show $f'(x) > 0$ for all $x > -1$.

Step 1: Differentiate term by term.

Step 2: Combine into a single fraction.

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

Why this approach works

The standard way to prove a function is increasing on an interval is to check its derivative. If $f'(x) \geq 0$ for all $x$ in the domain (and $f'(x) = 0$ only at isolated points), then $f$ is increasing. This is the Increasing Function Test — a direct consequence of the Mean Value Theorem.

Here, the domain is $x > -1$. The function is a combination of a log term and a rational term. The log term $\log(1+x)$ is itself increasing, but the subtraction of $\frac{2x}{2+x}$ could potentially reverse that. So we must check the net effect.

Step-by-step solution

1. Write the function and differentiate.

Let $f(x) = \log(1+x) - \frac{2x}{2+x}$, for $x > -1$.

Differentiate term by term:

• Derivative of $\log(1+x)$ is $\frac{1}{1+x}$.
• For $-\frac{2x}{2+x}$, use the quotient rule: $\frac{d}{dx}\left(\frac{2x}{2+x}\right) = \frac{(2)(2+x) - (2x)(1)}{(2+x)^2} = \frac{4+2x - 2x}{(2+x)^2} = \frac{4}{(2+x)^2}$.

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

2. Combine into a single fraction.

Get a common denominator: $(1+x)(2+x)^2$.

3. Simplify the numerator.

Expand $(2+x)^2 = x^2 + 4x + 4$.

Then numerator becomes:

So:

4. Analyse the sign of $f'(x)$ for $x > -1$.

• The numerator $x^2 \geq 0$ for all real $x$, and equals $0$ only at $x = 0$.
• Denominator: $(1+x) > 0$ for $x > -1$ (since $x > -1 \implies 1+x > 0$). Also $(2+x)^2 > 0$ for all $x \neq -2$, and since $x > -1$, it's certainly positive.

Thus for all $x > -1$, $x \neq 0$, we have $f'(x) > 0$. At $x = 0$, $f'(0) = 0$. …

Method: Proving a Function is Increasing Throughout Its Domain by Simplifying the Derivative

When a function's raw derivative looks like a difference of two terms (so its sign isn't obvious at a glance), the standard technique is to combine everything into a single fraction and simplify the numerator until its sign becomes self-evident — often collapsing to a perfect square or another manifestly non-negative expression.

Steps

Step 1: Differentiate each term separately

Differentiate any logarithmic or algebraic terms individually, using the quotient rule where needed for a fraction like $\dfrac{ax}{b+x}$.

Step 2: Combine all terms over a single common denominator

Rewrite $f'(x)$ as one fraction rather than a sum/difference of separate pieces — this is essential because the sign of a sum of terms with different denominators is not obvious, but the sign of a single fraction is much easier to analyse.

Step 3: Expand and simplify the numerator fully

Multiply out and collect like terms. Look for cancellation — well-designed textbook problems are often built so the numerator simplifies dramatically, frequently down to something like $x^2$ or another perfect square.

Step 4: Argue the sign of the numerator and denominator separately over the given domain …

Common Mistakes

Mistake 1: Concluding the function is not increasing because $f'(0)=0$

Why it's wrong: The Increasing Function Test only requires $f'(x) \ge 0$ throughout the interval, with equality allowed at isolated points — it does not require $f'(x) > 0$ everywhere without exception. Seeing $f'(0)=0$ and declaring the proof "fails" ignores that a single isolated zero doesn't break the increasing trend. Correct approach: check whether the zero occurs only at isolated points (not across a whole sub-interval); if so, the function is still increasing.

Mistake 2: Sign error while applying the quotient rule to $\frac{2x}{2+x}$ …