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Derivative of a sum of inverse tangents using tan^-1 a - tan^-1 b

This type of question, asked 8 times: 3 May 2023 S2, 16 May 2023 S2, 10 May 2024 S2, 23 Apr 2025 S1, 25 Apr 2025 S1, 13 Apr 2026 S2, 18 Apr 2026 S1, 18 Apr 2026 S2. All 8 questions below, oldest first.

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  1. Question 1

    3 May 2023 S2

    141
    If y=tan⁡−1 ⁣(3+2x2−3x)+tan⁡−1 ⁣(3x1+4x2)y=\tan^{-1}\!\left(\frac{3+2x}{2-3x}\right)+\tan^{-1}\!\left(\frac{3x}{1+4x^{2}}\right), then dydx\frac{dy}{dx} is equal to

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  2. Question 2

    16 May 2023 S2

    119
    If y=tan⁡−12+3x3−2x+tan⁡−14x1+5x2y = \tan^{-1}\frac{2+3x}{3-2x} + \tan^{-1}\frac{4x}{1+5x^2}, then dydx=\frac{dy}{dx} =

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  3. Question 3

    10 May 2024 S2

    120
    If y=tan⁡−1 ⁣(log⁡(e/x2)log⁡(ex2))+tan⁡−1 ⁣(4+2log⁡x1−8log⁡x)y=\tan^{-1}\!\left(\dfrac{\log(e/x^2)}{\log(ex^2)}\right)+\tan^{-1}\!\left(\dfrac{4+2\log x}{1-8\log x}\right), then dydx\dfrac{dy}{dx} is

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  4. Question 4

    23 Apr 2025 S1

    135
    If y=tan⁡−1[4sin⁡2xcos⁡2x−6sin⁡2x]y=\tan^{- 1}\left\lbrack \frac{4\sin2x}{\cos2x- 6\sin^{2}x} \right\rbrack, then dy dx\frac{dy}{\text{ }dx} at x=0x= 0 is

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  5. Question 5

    25 Apr 2025 S1

    112
    If y=tan⁡−1(4x1+5x2)+cot⁡−1(3−2x2+3x)y=\tan^{- 1}\left( \frac{4x}{1 + 5x^{2}} \right)+\cot^{- 1}\left( \frac{3 - 2x}{2 + 3x} \right), then dydx\frac{dy}{dx} is equal to

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  6. Question 6

    13 Apr 2026 S2

    66
    If y=tan⁡−1(1x2+x+1)+tan⁡−1(1x2+3x+3)+tan⁡−1(1x2+5x+7)+…y = \tan^{-1}\left(\dfrac{1}{x^2 + x + 1}\right) + \tan^{-1}\left(\dfrac{1}{x^2 + 3x + 3}\right) + \tan^{-1}\left(\dfrac{1}{x^2 + 5x + 7}\right) + \ldots upto nn terms, then y′(0)=y'(0) =

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  7. Question 7

    18 Apr 2026 S1

    69
    If y=tan⁡−1(log⁡(ex3)log⁡ex3)+tan⁡−1(log⁡(e4x3)log⁡(ex12)),x∈(e−13,e112)y = \tan^{-1}\left(\dfrac{\log\left(\frac{e}{x^3}\right)}{\log ex^3}\right) + \tan^{-1}\left(\dfrac{\log(e^4x^3)}{\log\left(\frac{e}{x^{12}}\right)}\right), x \in \left(e^{-\frac{1}{3}}, e^{\frac{1}{12}}\right) then dydx\dfrac{dy}{dx} is equal to...

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  8. Question 8

    18 Apr 2026 S2

    65
    For x>0x > 0 and (xlog⁡x)<1(x\log x) < 1, if y=cot⁡−1(x−log⁡xx2log⁡ex2+log⁡xx)y = \cot^{-1}\left(\dfrac{x - \log x^{x^2}}{\log e^{x^2} + \log x^x}\right), then dydx=…\dfrac{dy}{dx} = \ldots

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