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

Concept: Inverse Tangent Identity — Use the formula for $\tan^{-1} u + \tan^{-1} v$ to simplify the argument into a single term.

Let $y = \tan^{-1}\left(\dfrac{3a^2 x - x^3}{a^3 - 3ax^2}\right)$.

Notice that the numerator and denominator resemble $\tan 3\theta = \dfrac{3\tan\theta - \tan^3\theta}{1 - 3\tan^2\theta}$.

Set $\dfrac{x}{a} = \tan\theta$, so $x = a\tan\theta$. Then

$\frac{3a^2 x - x^3}{a^3 - 3ax^2} = \frac{3a^3\tan\theta - a^3\tan^3\theta}{a^3 - 3a^3\tan^2\theta} = \frac{3\tan\theta - \tan^3\theta}{1 - 3\tan^2\theta} = \tan 3\theta.$ …

We start with the function

and the condition $-\frac{1}{\sqrt{3}} < \frac{x}{a} < \frac{1}{\sqrt{3}}$.

The key insight is that the fraction inside the inverse tangent resembles the formula for $\tan 3\theta$ in terms of $\tan \theta$. Recall:

If we set $\tan\theta = \frac{x}{a}$, then

That’s exactly the argument of the inverse tangent. So

where $\theta = \tan^{-1}\left(\frac{x}{a}\right)$.

Now, the identity $\tan^{-1}(\tan \alpha) = \alpha$ holds only when $\alpha$ lies in the principal branch $(-\pi/2, \pi/2)$. Here $\alpha = 3\theta = 3\tan^{-1}(x/a)$. The given condition $-\frac{1}{\sqrt{3}} < \frac{x}{a} < \frac{1}{\sqrt{3}}$ ensures that $\tan^{-1}(x/a)$ lies between $-\pi/6$ and $\pi/6$, so $3\tan^{-1}(x/a)$ lies between $-\pi/2$ and $\pi/2$. Perfect — we are safely inside the principal range.

Thus, …

Method: Trigonometric Substitution to Exploit the $\tan 3\theta$ Identity

Whenever the argument of an inverse tangent has the specific algebraic shape $\dfrac{3a^2x-x^3}{a^3-3ax^2}$ (or, after factoring, $\dfrac{3(x/a)-(x/a)^3}{1-3(x/a)^2}$), recognise it as the triple-angle tangent formula in disguise — substituting $\tan\theta = x/a$ collapses the whole expression to a single angle.

Steps

Step 1: Set $\tan\theta = x/a$

This is the substitution to try whenever a $3(\cdot)-(\cdot)^3$ over $1-3(\cdot)^2$ pattern appears, since it matches

Step 2: Factor the given expression into the same form

Factor $a^3$ out of both the numerator and denominator of $\dfrac{3a^2x-x^3}{a^3-3ax^2}$ and divide through, so it becomes $\dfrac{3(x/a)-(x/a)^3}{1-3(x/a)^2} = \tan 3\theta$.

Step 3: Simplify $y=\tan^{-1}(\tan 3\theta)$, checking the branch …

Common Mistakes

Mistake 1: Not verifying that $3\theta$ stays within the principal branch

Why it's wrong: the identity $\tan^{-1}(\tan\alpha)=\alpha$ only holds when $\alpha\in(-\pi/2,\pi/2)$; the specific domain restriction given ($-1/\sqrt3<x/a<1/\sqrt3$) exists precisely to guarantee this for $3\theta$, and treating the simplification as automatically valid for any $x$ is a genuine error, not a technicality. Correct approach: translate the given domain into a bound on $\theta$, triple it, and confirm the result lands inside the principal branch before simplifying.

Mistake 2: Forgetting the factor of $\dfrac1a$ when differentiating $\tan^{-1}(x/a)$

Why it's wrong: by the chain rule, $\dfrac{d}{dx}\tan^{-1}\!\left(\dfrac xa\right) = \dfrac{1}{1+(x/a)^2}\cdot\dfrac1a$ — omitting the inner derivative $\dfrac1a$ (treating $x/a$ as if it were just $x$) gives an answer missing a factor of $a$. Correct approach: always differentiate the inner argument $x/a$ explicitly as its own chain-rule step. …