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Derivative of tan^-1 or sin^-1 of a double- or triple-angle form

This type of question, asked 4 times: 9 May 2023 S1, 9 May 2023 S2, 25 Apr 2025 S2, 15 May 2026 S2. All 4 questions below, oldest first.

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  1. Question 1

    9 May 2023 S1

    112
    If for x∈(0,14)x\in\left(0,\frac{1}{4}\right), the derivative of tan⁡−1 ⁣(6xx1−9x3)\tan^{-1}\!\left(\frac{6x\sqrt{x}}{1-9x^3}\right) is x⋅g(x)x\cdot g(x), then g(x)g(x) equals

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  2. Question 2

    9 May 2023 S2

    114
    If y=sin⁡−1(3x2−x32)y = \sin^{-1}\left(\frac{3x}{2} - \frac{x^3}{2}\right), then dydx\frac{dy}{dx} is equal to

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  3. Question 3

    25 Apr 2025 S2

    116
    If y=tan⁡−1(12x−64x31−48x2)y=\tan^{- 1}\left( \frac{12x- 64x^{3}}{1 - 48x^{2}} \right), then dydx=\frac{dy}{dx}=

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  4. Question 4

    15 May 2026 S2

    69
    If y=tan⁡−1(2x5/21−x5)y = \tan^{-1}\left(\dfrac{2x^{5/2}}{1 - x^5}\right), then dydx\dfrac{dy}{dx} is equal to ....

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