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Derivative of one inverse-trig expression with respect to another, simplified by a substitution

This type of question, asked 10 times: May 2021 S1, 10 May 2023 S2, 15 May 2023 S2, 11 May 2024 S1, 22 Apr 2025 S2, 11 Apr 2026 S2, 16 Apr 2026 S2, 17 Apr 2026 S2, 18 Apr 2026 S1, 14 May 2026 S2. All 11 questions below, oldest first.

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  1. Question 1

    May 2021 S1

    141
    Derivative of tan⁡−1 ⁣1+x2−1x\tan^{-1}\!\frac{\sqrt{1+x^2}-1}{x} w.r.t. sin⁡−1 ⁣2x1+x2\sin^{-1}\!\frac{2x}{1+x^2} is

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  2. Question 2

    10 May 2023 S2

    136
    Derivative of tan⁡−1 ⁣(1−x1+x)\tan^{-1}\!\left(\frac{1-x}{1+x}\right) w.r.t. cos⁡−1(4x3−3x)\cos^{-1}(4x^3-3x) is

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  3. Question 3

    15 May 2023 S2

    116
    The derivative of sin⁡−1(2x1−x2)\sin^{-1}(2x\sqrt{1-x^2}) w.r.t. sin⁡−1(3x−4x3)\sin^{-1}(3x-4x^3) is

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  4. Question 4

    11 May 2024 S1

    130
    Derivative of tan⁡−1 ⁣(1+x2−1−x21+x2+1−x2)\tan^{-1}\!\left(\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}\right) w.r.t. cos⁡−1(x2)\cos^{-1}(x^2) is

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  5. Question 5

    22 Apr 2025 S2

    133
    The derivative of tan⁡−1(1+x2−1x)\tan^{- 1}\left( \frac{\sqrt{1 +x^{2}}- 1}{x} \right) w.r.t. tan⁡−1(2x1−x21−2x2)\tan^{- 1}\left( \frac{2x\sqrt{1 -x^{2}}}{1 - 2x^{2}} \right) at x=0x= 0 is

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  6. Question 6

    11 Apr 2026 S2

    70
    The derivative of tan⁡−1(1+x2−1x)\tan^{-1}\left(\dfrac{\sqrt{1 + x^2} - 1}{x}\right) with respect to sin⁡−1(2x1+x2)\sin^{-1}\left(\dfrac{2x}{1 + x^2}\right) is...

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  7. Question 7

    16 Apr 2026 S2

    72
    If cot⁡[f(x)]=3x−x31−3x2\cot[f(x)] = \dfrac{3x - x^3}{1 - 3x^2} and sin⁡[g(x)]=1−x21+x2\sin[g(x)] = \dfrac{1 - x^2}{1 + x^2}, then lim⁡x→tf(x)−f(t)g(x)−g(t)=…\displaystyle\lim_{x \to t}\dfrac{f(x) - f(t)}{g(x) - g(t)} = \ldots

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  8. Question 8

    17 Apr 2026 S2

    71
    The derivative of tan⁡−1(1+x2−1x)\tan^{-1}\left(\dfrac{\sqrt{1 + x^2} - 1}{x}\right) with respect to tan⁡−1(x1−x2)\tan^{-1}\left(\dfrac{x}{\sqrt{1 - x^2}}\right) at x=12x = \dfrac{1}{2} is

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  9. Question 9

    17 Apr 2026 S2

    73
    If f(x)=cot⁡−1(3x−x31−3x2)f(x) = \cot^{-1}\left(\dfrac{3x - x^3}{1 - 3x^2}\right) and g(x)=cos⁡−1(1−x21+x2)g(x) = \cos^{-1}\left(\dfrac{1 - x^2}{1 + x^2}\right), then lim⁡x→af(x)−f(a)g(x)−g(a), (0<a<12)\lim_{x \to a}\dfrac{f(x) - f(a)}{g(x) - g(a)},\ \left(0 < a < \dfrac{1}{2}\right) is

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  10. Question 10

    18 Apr 2026 S1

    68
    Let t∈(0,1)t \in (0, 1) and α∈(0,π4)\alpha \in \left(0, \dfrac{\pi}{4}\right). If x=csc⁡−1(1+t22t),y=cot⁡−1(1−t2t)x = \mathrm{\csc}^{-1}\left(\dfrac{1 + t^2}{2t}\right), y = \cot^{-1}\left(\dfrac{\sqrt{1 - t^2}}{t}\right) and dydx=f(t)\dfrac{dy}{dx} = f(t), then the value of f(tan⁡α)f(\tan\alpha) is

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  11. Question 11

    14 May 2026 S2

    67
    If u=tan⁡−1(1+x2−1x)u = \tan^{-1}\left(\dfrac{\sqrt{1 + x^2} - 1}{x}\right) and v=tan⁡−1(2x1−x21−2x2)v = \tan^{-1}\left(\dfrac{2x\sqrt{1 - x^2}}{1 - 2x^2}\right) then dudv=…\dfrac{du}{dv} = \ldots

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