Q.Differentiate w.r.t. $x$: $\sec^{-1}\left(\dfrac{1}{4x^3 - 3x}\right),\ 0 < x < \dfrac{1}{\sqrt{2}}$.

Concept: Chain rule via a trigonometric substitution.

Put $x=\cos\theta$ with $\theta=\cos^{-1}x$. For $0<x<\frac{1}{\sqrt2}$, $\theta\in(\frac{\pi}{4},\frac{\pi}{2})$. The triple-angle identity gives $4x^3-3x=4\cos^3\theta-3\cos\theta=\cos 3\theta$, so

Here $3\theta\in(\frac{3\pi}{4},\frac{3\pi}{2})$, which crosses $\pi$ at $\theta=\frac{\pi}{3}$ (i.e. $x=\frac12$), so $\sec^{-1}(\sec 3\theta)$ folds. Using $\frac{dx}{d\theta}=-\sin\theta$ and $\sin\theta=\sqrt{1-x^2}$:

• $0<x<\frac12$ (so $3\theta>\pi$): $y=2\pi-3\theta$, giving $\frac{dy}{dx}=\frac{-3}{-\sin\theta}=\frac{3}{\sqrt{1-x^2}}$. …

Set up

Let

The form $4x^3-3x$ is the signal to substitute $x=\cos\theta$, because $4\cos^3\theta-3\cos\theta=\cos 3\theta$.

Substitute

Take $\theta=\cos^{-1}x$. Since $x\in(0,\frac{1}{\sqrt2})$, $\theta\in(\frac{\pi}{4},\frac{\pi}{2})$. Then

Fold into the principal range

$\sec^{-1}$ returns values in $[0,\pi]\setminus\{\frac{\pi}{2}\}$. As $\theta$ runs over $(\frac{\pi}{4},\frac{\pi}{2})$, $3\theta$ runs over $(\frac{3\pi}{4},\frac{3\pi}{2})$ and passes through $\pi$ at $\theta=\frac{\pi}{3}$, i.e. $x=\frac12$:

• If $3\theta\in(\frac{3\pi}{4},\pi)$, i.e. $\theta\in(\frac{\pi}{4},\frac{\pi}{3})$, i.e. $x\in(\frac12,\frac{1}{\sqrt2})$: this is already in range, so $y=3\theta$.
• If $3\theta\in(\pi,\frac{3\pi}{2})$, i.e. $\theta\in(\frac{\pi}{3},\frac{\pi}{2})$, i.e. $x\in(0,\frac12)$: use $\sec 3\theta=\sec(2\pi-3\theta)$ with $2\pi-3\theta\in(\frac{\pi}{2},\pi)$, so $y=2\pi-3\theta$. …

Method: Trigonometric Substitution to Exploit a Triple-Angle Identity, With Careful Branch Folding

Whenever an expression like $4x^3-3x$ appears inside an inverse trig function, recognise it as the triple-angle cosine formula in disguise — but because inverse trig functions have a restricted principal range, simplifying "$\sec^{-1}(\sec 3\theta)=3\theta$" is only valid where $3\theta$ actually lies in that range; elsewhere you must fold the angle back in.

Steps

Step 1: Substitute $x=\cos\theta$

This is the signal substitution whenever you see $4x^3-3x$, because $4\cos^3\theta - 3\cos\theta = \cos 3\theta$ (the triple-angle identity).

Step 2: Rewrite the given expression using the identity

Step 3: Determine the range of $3\theta$ implied by the given domain of $x$

Convert the given interval for $x$ into the corresponding interval for $\theta=\cos^{-1}x$, then triple it. Check whether this range for $3\theta$ falls entirely inside $\sec^{-1}$'s principal range $[0,\pi]\setminus\{\pi/2\}$.

Step 4: Fold any portion outside the principal range back in …

Common Mistakes

Mistake 1: Assuming $\sec^{-1}(\sec 3\theta)=3\theta$ holds across the entire given domain

Why it's wrong: $\sec^{-1}$ only undoes $\sec$ cleanly when its argument already lies in the principal range $[0,\pi]\setminus\{\pi/2\}$ — for part of this problem's domain, $3\theta$ falls outside that range, so the naive substitution gives an unfolded, wrong expression for $y$ on that portion. Correct approach: always check whether $3\theta$ stays inside the principal range across the whole given domain, not just at one sample value.

Mistake 2: Treating the function as differentiable by a single formula across the whole domain …