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

Concept: Increasing Function Test — A function $f$ is increasing on an interval if $f'(x) > 0$ for all $x$ in that interval.

Step 1: Differentiate $f(x) = \tan^{-1}(\sin x + \cos x)$.

Using the chain rule:

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

The core idea here is the Increasing Function Test from calculus: if $f'(x) > 0$ for all $x$ in an open interval, then $f$ is strictly increasing on that interval. So the problem reduces to checking the sign of $f'(x)$ on $\left(0, \frac{\pi}{4}\right)$.

Why does this work? Because the derivative tells us the instantaneous rate of change. A positive derivative means the function's output rises as $x$ increases — exactly what "increasing" means.

Let's compute $f'(x)$ carefully.

1. Differentiate the outer function. $f(x) = \tan^{-1}(u)$ where $u = \sin x + \cos x$. The derivative of $\tan^{-1} u$ is $\frac{1}{1+u^2} \cdot \frac{du}{dx}$. So:

2. Analyze the sign.

   The denominator $1 + (\sin x + \cos x)^2$ is always positive — it's $1$ plus a square. So the sign of $f'(x)$ is entirely determined by the numerator $(\cos x - \sin x)$.

3. Compare $\cos x$ and $\sin x$ on $(0, \pi/4)$.

   On the interval $(0, \pi/4)$, we know:

   • 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.
   • For any $x$ strictly between $0$ and $\pi/4$, $\cos x$ is decreasing from $1$ to $\frac{\sqrt{2}}{2}$, while $\sin x$ is increasing from $0$ to $\frac{\sqrt{2}}{2}$. Since $\cos x$ starts larger and both change continuously, $\cos x > \sin x$ throughout $(0, \pi/4)$.

   Therefore $\cos x - \sin x > 0$ for all $x \in (0, \pi/4)$. …

Method: Proving a Function Is Increasing on an Interval

The general technique for proving — not just locating — that a given function is increasing throughout a specified interval.

Steps

Step 1: Differentiate using the chain rule for the composite function

For $f(x) = g(h(x))$ (here $g = \tan^{-1}$ and $h(x) = \sin x + \cos x$), the derivative is

Being a composite of an inverse-trig outer function with a trig inner function doesn't change the method — the chain rule still applies exactly as it would for any other composite.

Step 2: Isolate the part of $f'(x)$ whose sign is actually in question

The derivative of $\tan^{-1}(u)$ contributes a factor of $\frac{1}{1+u^2}$, which is always strictly positive for any real $u$, so this factor can never flip the sign of $f'(x)$. The entire sign question reduces to the sign of the remaining factor, $h'(x)$.

Step 3: Prove that factor is positive on exactly the given interval …

Common Mistakes

Mistake 1: Assuming $\tan^{-1}(u)$ is automatically increasing without checking the inner function

Why it's wrong: $\tan^{-1}(u)$ is indeed an increasing function of $u$, but $f(x)=\tan^{-1}(\sin x+\cos x)$ is a composite function — its monotonicity in $x$ depends on the chain-rule product $g'(h(x))\cdot h'(x)$, not on the outer function's behavior alone. If the inner function $h(x)=\sin x+\cos x$ were decreasing somewhere, $f$ would be decreasing there too, despite $\tan^{-1}$ itself always increasing in its own variable. Correct approach: always differentiate the full composite with the chain rule and check the sign of the actual derivative, never assume monotonicity transfers automatically from the outer function.

Mistake 2: Testing the inequality $\cos x > \sin x$ at only a single point instead of proving it for the whole interval …