Q.Show that $f(x) = \tan^{-1}(\sin x + \cos x)$ is an increasing function in $\left(0, \dfrac{\pi}{4}\right)$.

Concept: Monotonicity of Trigonometric Functions — we check the sign of $f'(x)$ in the given interval.

Step 1: Differentiate $f(x)$.

Step 2: The denominator $1 + (\sin x + \cos x)^2 > 0$ for all $x$. So the sign of $f'(x)$ depends only on $\cos x - \sin x$. …

The core idea here is monotonicity of composite functions. If an outer function is strictly increasing, then the composite inherits the monotonicity of the inner function. This is a powerful shortcut — instead of differentiating the whole mess, we can focus on the simpler part.

Here, $f(x) = \tan^{-1}(\sin x + \cos x)$. The outer function $\tan^{-1}$ (or $\arctan$) is strictly increasing on $\mathbb{R}$ — its derivative $\frac{1}{1+x^2}$ is always positive. So $f$ will increase exactly when its inner function $g(x) = \sin x + \cos x$ increases.

So the problem reduces to: Show $g(x) = \sin x + \cos x$ is increasing on $(0, \pi/4)$.

Let's work through it.

1. Find the derivative of $g$.

   $g'(x) = \cos x - \sin x$.

   This is straightforward — derivative of $\sin x$ is $\cos x$, derivative of $\cos x$ is $-\sin x$.

2. Analyse the sign of $g'(x)$ on $(0, \pi/4)$.

   On this interval, both $\cos x$ and $\sin x$ are positive. But which is larger?

   At $x=0$: $\cos 0 = 1$, $\sin 0 = 0$, so $\cos x > \sin x$.

   At $x=\pi/4$: $\cos(\pi/4) = \sin(\pi/4) = \frac{\sqrt{2}}{2}$, so they are equal.

   Since $\cos x$ decreases and $\sin x$ increases on $(0, \pi/2)$, the difference $\cos x - \sin x$ is positive for $x < \pi/4$ and zero at $x = \pi/4$. …

Method: Monotonicity of a Composite Function via the Sign of the Derivative

This method proves a function is increasing or decreasing on an interval by differentiating once and analysing the sign of a single, simpler factor — the standard approach whenever the function is built from an outer function whose own derivative is always positive, wrapped around a simpler inner expression.

Steps

Step 1: Differentiate using the Chain Rule

Write the function as a composition, $f(x) = h(g(x))$, and differentiate:

For $h(u) = \tan^{-1} u$, recall $h'(u) = \dfrac{1}{1+u^2}$ — this factor is always positive, no matter what $u$ is.

Step 2: Isolate the factor that actually controls the sign

Since $h'(g(x)) > 0$ always, the sign of $f'(x)$ depends entirely on the sign of $g'(x)$ (the derivative of the inner expression). This is the key simplification: you don't need to analyse the whole messy expression for $f'(x)$ — only the simpler factor.

Step 3: Determine the sign of that factor on the given interval …

Common Mistakes

Mistake 1: Differentiating $f(x)=\tan^{-1}(\sin x+\cos x)$ directly and losing track of the sign

A student who differentiates the whole composite in one shot gets $f'(x)=\dfrac{\cos x-\sin x}{1+(\sin x+\cos x)^2}$, and then sometimes stops at the numerator without checking whether the denominator could ever be zero or negative. Why it's wrong: skipping the denominator check leaves the sign argument incomplete, even though $1+(\text{anything})^2>0$ always holds. Correct approach: explicitly state the denominator is $1$ plus a square, hence always positive, so the sign of $f'(x)$ depends purely on $\cos x-\sin x$ — that one line is what actually proves the claim.

Mistake 2: Asserting $\cos x>\sin x$ on $\left(0,\tfrac{\pi}{4}\right)$ without justifying it …