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Derivative of a trig function of an inverse-trig function (convert to algebra first)

This type of question, asked 10 times: May 2022 S1, 2 May 2023 S2, 3 May 2023 S1, 10 May 2023 S2, 16 May 2023 S1, 21 Apr 2025 S2, 13 Apr 2026 S2, 19 Apr 2026 S1, 20 Apr 2026 S2, 13 May 2026 S2. All 10 questions below, oldest first.

Try each one first, then open its answer.

  1. Question 1

    May 2022 S1

    120
    If y=sin⁡ ⁣(2tan⁡−1 ⁣1+x1−x)y = \sin\!\left(2\tan^{-1}\!\sqrt{\frac{1+x}{1-x}}\right) then dydx\frac{dy}{dx} is equal to

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  2. Question 2

    2 May 2023 S2

    145
    Let f(θ)=sin⁡ ⁣(tan⁡−1 ⁣sin⁡θcos⁡2θ)f(\theta) = \sin\!\left(\tan^{-1}\!\frac{\sin\theta}{\sqrt{\cos2\theta}}\right) where −π4<θ<π4-\frac{\pi}{4}<\theta<\frac{\pi}{4}, then dd(tan⁡θ)(f(θ))\frac{d}{d(\tan\theta)}(f(\theta)) is

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  3. Question 3

    3 May 2023 S1

    126
    If y=sec⁡(tan⁡−1x)y=\sec(\tan^{-1}x), then dydx\frac{dy}{dx} at x=1x=1 is equal to

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  4. Question 4

    10 May 2023 S2

    125
    If y=sin⁡2 ⁣(cot⁡−1 ⁣1+x1−x)y = \sin^2\!\left(\cot^{-1}\!\sqrt{\frac{1+x}{1-x}}\right), then dydx\frac{dy}{dx} has the value

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  5. Question 5

    16 May 2023 S1

    131
    If y=sec⁡(tan⁡−1x)y = \sec(\tan^{-1}x), then dydx\frac{dy}{dx} at x=1x=1 is equal to

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  6. Question 6

    21 Apr 2025 S2

    114
    If y=sin⁡2(cot⁡−1(1−x1+x))y=\sin^{2}\left( \cot^{- 1}\left( \sqrt{\frac{1 -x}{1 +x}} \right) \right), then dy dx=\frac{dy}{\text{ }dx}=

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  7. Question 7

    13 Apr 2026 S2

    71
    ddx[sin⁡2{cot⁡−11−x1+x}]=…\dfrac{d}{dx}\left[\sin^2\left\{\cot^{-1}\sqrt{\dfrac{1 - x}{1 + x}}\right\}\right] = \ldots

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  8. Question 8

    19 Apr 2026 S1

    71
    If y=sin⁡(2sin⁡−1x)y = \sin(2\sin^{-1}x) then dydx=\dfrac{dy}{dx} =..

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  9. Question 9

    20 Apr 2026 S2

    71
    If y=cos⁡2[cot⁡−1(1−x1+x)]y = \cos^2\left[\cot^{-1}\left(\sqrt{\dfrac{1 - x}{1 + x}}\right)\right] then dydx=\dfrac{dy}{dx} = ........

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  10. Question 10

    13 May 2026 S2

    73
    If y=sin⁡(2sin⁡−1x)y = \sin(2\sin^{-1}x), then dydx=\dfrac{dy}{dx} = ______

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