Q.Find $\frac{dy}{dx}$ in the following: $y = \sec^{-1} \left(\frac{1}{2x^2-1}\right), 0 < x < \frac{1}{\sqrt{2}}$

Step 1: Let $x = \cos\theta$. Since $0 < x < \tfrac{1}{\sqrt2}$, $\theta \in \left(\tfrac{\pi}{4},\tfrac{\pi}{2}\right)$. Then

Step 2: Since $\theta \in \left(\tfrac{\pi}{4},\tfrac{\pi}{2}\right)$, $2\theta \in \left(\tfrac{\pi}{2},\pi\right)$, which lies in the principal range $[0,\pi]\setminus\{\tfrac{\pi}{2}\}$ of $\sec^{-1}$. So $y = 2\theta = 2\cos^{-1}x$ directly. …

Recognising the Identity

The expression $2x^2 - 1$ is the double-angle formula $\cos 2\theta = 2\cos^2\theta - 1$ in terms of $x = \cos\theta$, so $\dfrac{1}{2x^2-1}$ becomes $\sec 2\theta$ — the natural argument for $\sec^{-1}$.

Step-by-Step Solution

1. Substitute $x = \cos\theta$.

Since cosine is a decreasing bijection from $[0,\pi]$ onto $[-1,1]$, let $\theta = \cos^{-1}x$. With $\cos\tfrac{\pi}{4} = \tfrac{1}{\sqrt2}$ and $\cos\tfrac{\pi}{2} = 0$, the given domain $0 < x < \tfrac{1}{\sqrt2}$ corresponds to:

2. Rewrite the argument using the double-angle identity.

(Check of sign: $x \in \left(0, \tfrac{1}{\sqrt2}\right) \Rightarrow x^2 \in \left(0, \tfrac12\right) \Rightarrow 2x^2 - 1 \in (-1, 0)$, so $\sec 2\theta$ is negative here — consistent with $2\theta \in \left(\tfrac{\pi}{2}, \pi\right)$ found below.)

So:

3. Check the range of $2\theta$.

Since $\theta \in \left(\tfrac{\pi}{4}, \tfrac{\pi}{2}\right)$:

$2\theta \in \left(\frac{\pi}{2}, \pi\right),$ …

Method: Simplify Inverse-Trig Compositions by Trigonometric Substitution (Secant/Cosine form)

This method applies when the argument inside an inverse-secant function is built from $x$ using a double-angle cosine pattern such as $2x^2-1$.

Steps

Step 1: Recognise the double-angle pattern

The expression $2x^2-1$ matches $2\cos^2\theta - 1 = \cos 2\theta$. This signals the substitution $x=\cos\theta$, restricting $\theta$ to $[0,\pi]$ (the principal domain of $\cos^{-1}$).

Step 2: Substitute and simplify the argument

so the original expression becomes $\sec^{-1}(\sec 2\theta)$.

Step 3: Convert the domain of $x$ into a range for $2\theta$, and check the principal range …

Common Mistakes

Mistake 1: Leaving the answer as a simplified $y$ instead of computing $\dfrac{dy}{dx}$

Why it's wrong: the question explicitly asks to find $\dfrac{dy}{dx}$; reducing $y$ to $2\cos^{-1}x$ is only the setup. Correct approach: always finish with $\dfrac{dy}{dx} = \dfrac{d}{dx}(2\cos^{-1}x) = -\dfrac{2}{\sqrt{1-x^2}}$.

Mistake 2: Confusing the domain of $\sec^{-1}x$ with that of $\cos^{-1}x$

Why it's wrong: $\sec^{-1}x$ is only defined for $|x|\ge 1$, while the substitution variable in $x=\cos\theta$ ranges over $[-1,1]$ — mixing these up leads to checking the wrong inequality when verifying the argument is a valid input for $\sec^{-1}$. Correct approach: keep track of which quantity you mean at each stage — the original problem's $x\in(0,1/\sqrt2)$ versus the argument $\frac{1}{2x^2-1}$ that must satisfy $|{\cdot}|\ge 1$. …