Q.The derivative of $\cos^{-1}(2x^2 - 1)$ w.r.t. $\cos^{-1} x$ is
(A) $2$
(B) $\dfrac{-1}{2\sqrt{1 - x^2}}$
(C) $\dfrac{2}{x}$
(D) $1 - x^2$

Concept: Second derivative / derivative of inverse trigonometric functions using substitution.

Let $u = \cos^{-1}(2x^2 - 1)$ and $v = \cos^{-1} x$. We need $\frac{du}{dv}$.

Step 1: Put $x = \cos \theta$, so $\theta = \cos^{-1} x = v$.

Then $2x^2 - 1 = 2\cos^2\theta - 1 = \cos 2\theta$. …

The key insight here is that we are not differentiating with respect to $x$ directly. Instead, we are finding the rate of change of one function relative to another — a parametric derivative. The standard trick: if $u = \cos^{-1}(2x^2 - 1)$ and $v = \cos^{-1} x$, then $\frac{du}{dv} = \frac{du/dx}{dv/dx}$.

But there’s a deeper layer. The expression $2x^2 - 1$ is a classic double-angle disguise for cosine: $\cos(2\theta) = 2\cos^2\theta - 1$. So if we let $x = \cos\theta$, then $\cos^{-1} x = \theta$, and $\cos^{-1}(2x^2 - 1) = \cos^{-1}(\cos 2\theta)$. That substitution simplifies everything — but we must be careful with the range of inverse cosine.

Let’s work through it step by step.

1. Set up the parametric form.

   Let $v = \cos^{-1} x$ and $u = \cos^{-1}(2x^2 - 1)$. We need $\frac{du}{dv}$.

2. Differentiate each with respect to $x$.

   For $v$:

For $u$: use the chain rule.

The derivative of $2x^2 - 1$ is $4x$. So

3. Simplify the square root. Compute $(2x^2 - 1)^2 = 4x^4 - 4x^2 + 1$. Then

So

The absolute value matters — but we’ll handle it in a moment.

4. Form the ratio $\frac{du}{dv}$.

The negatives cancel. Also $\sqrt{1 - x^2}$ cancels top and bottom. We get:

5. Interpret the result. The expression $\frac{2x}{|x|}$ equals $2$ when $x > 0$, and $-2$ when $x < 0$. But the problem likely expects a constant answer from the given options — and only $2$ appears. Why? …

Method: Differentiating One Function w.r.t. Another via Trigonometric Substitution

When a problem asks for $\dfrac{du}{dv}$ where both $u$ and $v$ are (possibly composite) inverse trigonometric expressions in $x$, don't differentiate blindly with the chain rule term by term — first look for a trigonometric identity that collapses the composite expression into a simple multiple of the other variable.

Steps

Step 1: Identify the inner substitution suggested by the expression

Scan the argument of the inverse function for a recognisable double/triple-angle pattern — e.g. $2x^2-1$ is the cosine double-angle form $2\cos^2\theta-1=\cos2\theta$; $2x\sqrt{1-x^2}$ is $\sin2\theta$; $3x-4x^3$ is $\sin3\theta$; $\dfrac{2x}{1-x^2}$ is $\tan2\theta$. Choose $x=\cos\theta$ (or $\sin\theta$, $\tan\theta$) to match the 'outer' inverse function already present in the other variable, $v$.

Step 2: Rewrite both $u$ and $v$ purely in terms of $\theta$

With $v=\cos^{-1}x=\theta$ and the identity from Step 1, $u$ collapses to $\cos^{-1}(\cos k\theta)$ for some multiple $k$. Restrict $\theta$ to the range where $\cos^{-1}(\cos k\theta)=k\theta$ holds directly (i.e. $k\theta$ inside $[0,\pi]$) — this is exactly why these problems are normally stated (or assumed) for a restricted domain of $x$.

Step 3: Differentiate the resulting linear relation …

Common Mistakes

Mistake 1: Simplifying $\sqrt{4x^2(1-x^2)}$ as $2x\sqrt{1-x^2}$ instead of $2|x|\sqrt{1-x^2}$

Why it's wrong: $\sqrt{x^2}=|x|$, not $x$ — dropping the absolute value silently assumes $x>0$, which may not be justified by the question's domain. Correct approach: always write $\sqrt{x^2}=|x|$ first, then resolve the sign using the actual restricted domain the problem implies (here, the principal range that makes $\cos^{-1}(\cos2\theta)=2\theta$ forces $x\in(0,1)$, which is what finally makes the answer a clean constant).

Mistake 2: Substituting $x=\cos\theta$ but forgetting to restrict $\theta$'s range …