Q.Find $\frac{dy}{dx}$ in the following: $y = \tan^{-1} \left(\frac{3x-x^3}{1-3x^2}\right), -\frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}}$

Concept: Inverse Tangent Identity — the expression matches the tangent triple-angle formula, letting the inverse tangent be simplified before differentiating.

Step 1: Let $x = \tan\theta$. Since $\tan 3\theta = \dfrac{3\tan\theta - \tan^3\theta}{1-3\tan^2\theta}$,

$\frac{3x-x^3}{1-3x^2} = \tan 3\theta, \qquad y = \tan^{-1}(\tan 3\theta).$ …

Recognising the Identity

A cubic numerator over a quadratic denominator, with these specific coefficients ($3x - x^3$ over $1 - 3x^2$), is the signature of the tangent triple-angle formula:

Setting $x = \tan\theta$ turns the given expression into $\tan 3\theta$, but $\tan^{-1}(\tan 3\theta)$ only equals $3\theta$ when $3\theta$ lies in the principal range of $\tan^{-1}$ — the given domain on $x$ is exactly what guarantees this.

Step-by-Step Solution

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

Let $\theta = \tan^{-1}x$, so $\theta \in \left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right)$. Then:

2. Use the given domain to pin down the range of $3\theta$.

We are told $-\tfrac{1}{\sqrt3} < x < \tfrac{1}{\sqrt3}$. Since $\tan\tfrac{\pi}{6} = \tfrac{1}{\sqrt3}$ and $\tan$ is increasing on $\left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right)$:

3. Simplify using the principal range of $\tan^{-1}$.

Since $3\theta$ already lies inside $\left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right)$, the principal range of $\tan^{-1}$, no adjustment by $\pi$ is needed: …

Method: Recognising the Triple-Angle Pattern in an Inverse-Tangent Expression

Some inverse-tangent expressions hide the triple-angle tangent identity rather than the double-angle one — the giveaway is a cubic numerator paired with a quadratic denominator with alternating signs.

Steps

Step 1: Compare the expression's shape against the triple-angle formula

A numerator of the form $3(\cdot)-(\cdot)^3$ over a denominator $1-3(\cdot)^2$ is this identity, not the double-angle one ($\frac{2x}{1-x^2}$).

Step 2: Substitute $x=\tan\theta$

This turns the fraction into $\tan 3\theta$ exactly, so the expression becomes $y=\tan^{-1}(\tan 3\theta)$.

Step 3: Translate the given $x$-domain into a $\theta$-domain, then a $3\theta$-domain

Since $\tan\theta$ is strictly increasing on $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$, an inequality on $x=\tan\theta$ converts directly to the same inequality on $\theta$; multiplying through by $3$ gives the range of $3\theta$.

Step 4: Check $3\theta$ against $\tan^{-1}$'s principal range before cancelling …

Common Mistakes

Mistake 1: Confusing the triple-angle pattern with the double-angle one

Because both patterns involve $x=\tan\theta$, it's easy to misapply the double-angle formula $\frac{2x}{1-x^2}$ to an expression that's actually the triple-angle form $\frac{3x-x^3}{1-3x^2}$. Why it's wrong: using the wrong identity gives an entirely different (and incorrect) angle. Correct approach: check the powers carefully — a cubic numerator over a quadratic denominator with a factor of $3$ signals the triple-angle identity, $\tan 3\theta$, not the double-angle one.

Mistake 2: Skipping the range check because the answer "looks clean" …