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NCERT Exemplar · Q42

Q.Differentiate w.r.t. xx: tan⁡−1(3a2x−x3a3−3ax2), −13<xa<13\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}}.

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The given expression simplifies to 3tan⁡−1(xa)3\tan^{-1}\left(\frac{x}{a}\right) using the inverse tangent identity for triple angles, so its derivative is 3aa2+x2\frac{3a}{a^2 + x^2}.

We start with the function

y=tan⁡−1(3a2x−x3a3−3ax2)y = \tan^{-1}\left(\frac{3a^2 x - x^3}{a^3 - 3a x^2}\right)

and the condition −13<xa<13-\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⁡3θ\tan 3\theta in terms of tan⁡θ\tan \theta. Recall:

tan⁡3θ=3tan⁡θ−tan⁡3θ1−3tan⁡2θ\tan 3\theta = \frac{3\tan\theta - \tan^3\theta}{1 - 3\tan^2\theta}

If we set tan⁡θ=xa\tan\theta = \frac{x}{a}, then

tan⁡3θ=3(xa)−(xa)31−3(xa)2=3xa−x3a31−3x2a2=3a2x−x3a3a2−3x2a2=3a2x−x3a3−3ax2\tan 3\theta = \frac{3\left(\frac{x}{a}\right) - \left(\frac{x}{a}\right)^3}{1 - 3\left(\frac{x}{a}\right)^2} = \frac{\frac{3x}{a} - \frac{x^3}{a^3}}{1 - \frac{3x^2}{a^2}} = \frac{\frac{3a^2 x - x^3}{a^3}}{\frac{a^2 - 3x^2}{a^2}} = \frac{3a^2 x - x^3}{a^3 - 3a x^2}

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

y=tan⁡−1(tan⁡(3θ))y = \tan^{-1}\big(\tan(3\theta)\big)

where θ=tan⁡−1(xa)\theta = \tan^{-1}\left(\frac{x}{a}\right).

Now, the identity tan⁡−1(tan⁡α)=α\tan^{-1}(\tan \alpha) = \alpha holds only when α\alpha lies in the principal branch (−π/2,π/2)(-\pi/2, \pi/2). Here α=3θ=3tan⁡−1(x/a)\alpha = 3\theta = 3\tan^{-1}(x/a). The given condition −13<xa<13-\frac{1}{\sqrt{3}} < \frac{x}{a} < \frac{1}{\sqrt{3}} ensures that tan⁡−1(x/a)\tan^{-1}(x/a) lies between −π/6-\pi/6 and π/6\pi/6, so 3tan⁡−1(x/a)3\tan^{-1}(x/a) lies between −π/2-\pi/2 and π/2\pi/2. Perfect — we are safely inside the principal range.

Watch out

A common mistake is to forget the range condition. Without it, tan⁡−1(tan⁡3θ)\tan^{-1}(\tan 3\theta) might equal 3θ−π3\theta - \pi or 3θ+π3\theta + \pi, changing the derivative. Always check the interval.

Thus, …

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