Q.Show that $f(x) = 2x + \cot^{-1}x + \log\left(\sqrt{1+x^2} - x\right)$ is increasing in $\mathbb{R}$.

Key idea: Show $f'(x)\ge 0$ for every real $x$; then $f$ is increasing on $\mathbb{R}$.

Differentiate term by term:

For the log term, let $u=\sqrt{1+x^2}-x$. Then

so $\dfrac{u'}{u}=-\dfrac{1}{\sqrt{1+x^2}}$ (the factor $\sqrt{1+x^2}-x$ cancels).

Therefore

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

The idea

A differentiable function is increasing on an interval when its derivative never goes negative there. So the whole task is: compute $f'(x)$ and show $f'(x)\ge0$ for every real $x$.

Step 1 — differentiate the easy terms

• $\dfrac{d}{dx}(2x)=2$
• $\dfrac{d}{dx}\cot^{-1}x=-\dfrac{1}{1+x^2}$

Step 2 — the logarithmic term (the one to be careful with)

Let $u=\sqrt{1+x^2}-x$. Then

So

Note the $\sqrt{1+x^2}$ stays in the denominator — the derivative of this term is $-\dfrac{1}{\sqrt{1+x^2}}$, not $-1$.

Step 3 — assemble $f'(x)$

Step 4 — show it is non-negative …

Method: Proving a Function Is Increasing (or Decreasing) Everywhere Using the Sign of $f'(x)$

This is the standard approach whenever a function combines several terms (polynomial, inverse trig, logarithmic, etc.) and you must show $f$ is increasing (or decreasing) on its whole domain, per the Increasing/Decreasing Function Test: $f'(x) \ge 0$ everywhere $\Rightarrow f$ is increasing.

Steps

Step 1: Differentiate $f(x)$ term by term.

Handle each type of term with its own rule — polynomial terms directly, inverse trig terms with their standard derivative formulas, and any logarithmic term with the chain rule:

Step 2: Simplify any composite term carefully — watch for cancellation.

When a logarithm's argument $u$ involves a square root minus (or plus) $x$, computing $u'$ often produces a factor that is a multiple of $u$ itself, which cancels against the denominator. Don't skip writing this out — a common error is to lose the surviving $\sqrt{\cdots}$ term in the denominator by cancelling too aggressively.

Step 3: Combine all pieces into a single expression for $f'(x)$, then introduce a substitution to reveal its sign. …

Common Mistakes

Mistake 1: Mishandling the derivative of the log term

Students often differentiate $\log\left(\sqrt{1+x^2}-x\right)$ by treating it as though it were just $\log\sqrt{1+x^2}$, dropping the $-x$ inside, or fail to notice that the factor $\sqrt{1+x^2}-x$ cancels between numerator and denominator. Why it's wrong: skipping the cancellation leaves a messy, unsimplified expression that's nearly impossible to sign-analyze; the correct simplification collapses it neatly to $-\dfrac{1}{\sqrt{1+x^2}}$. Correct approach: differentiate the full inner expression $u=\sqrt{1+x^2}-x$ first, then form $u'/u$ and simplify algebraically before moving on.

Mistake 2: Sign error on $\cot^{-1}x$ …