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Derivative at x = 0 (or 1) of a long product by logarithmic differentiation

This type of question, asked 9 times: 3 May 2023 S2, 4 May 2023 S1, 11 May 2023 S1, 15 May 2023 S1, 11 May 2024 S2, 12 May 2024 S1, 13 May 2024 S2, 14 May 2024 S1, 17 Apr 2026 S1. All 9 questions below, oldest first.

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  1. Question 1

    3 May 2023 S2

    131
    If y=[(x+1)(2x+1)(3x+1)⋯(nx+1)]4y=[(x+1)(2x+1)(3x+1)\cdots(nx+1)]^{4} then dydx\frac{dy}{dx} at x=0x=0 is

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  2. Question 2

    4 May 2023 S1

    144
    If y=(x+1)(4x+1)(9x+1)⋯(n2x+1)2y=(x+1)(4x+1)(9x+1)\cdots(n^2x+1)^2, then dydx\frac{dy}{dx} at x=0x=0 is

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  3. Question 3

    11 May 2023 S1

    110
    If f(x)=(1+x)(1+x2)(1+x4)(1+x8)f(x) = (1+x)(1+x^2)(1+x^4)(1+x^8), then f′(1)=f'(1) =

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  4. Question 4

    15 May 2023 S1

    101
    If y=[(x+1)(2x+1)(3x+1)⋯(nx+1)]2y = [(x+1)(2x+1)(3x+1)\cdots(nx+1)]^{2}, then dydx\frac{dy}{dx} at x=0x=0 is

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  5. Question 5

    11 May 2024 S2

    133
    If y=[(x+1)(2x+1)(3x+1)⋯(nx+1)]3/2y=[(x+1)(2x+1)(3x+1)\cdots(nx+1)]^{3/2}, then dydx\frac{dy}{dx} at x=0x=0 is

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  6. Question 6

    12 May 2024 S1

    143
    If y=(1+x)(1+x2)(1+x4)⋯(1+x2n)y=(1+x)(1+x^{2})(1+x^{4})\cdots(1+x^{2^{n}}) then the value of dydx\frac{dy}{dx} at x=0x=0 is

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  7. Question 7

    13 May 2024 S2

    144
    y=(1+3x)1/3(1+4x)1/4(1+5x)1/5(1+7x)1/7(1+8x)1/8y=\frac{(1+3x)^{1/3}(1+4x)^{1/4}(1+5x)^{1/5}}{(1+7x)^{1/7}(1+8x)^{1/8}}. Then dydx\frac{dy}{dx} at x=0x=0 is:

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  8. Question 8

    14 May 2024 S1

    134
    If y=[(x+1)(2x+1)⋯(nx+1)]ny=[(x+1)(2x+1)\cdots(nx+1)]^n, then dydx\dfrac{dy}{dx} at x=0x=0 is

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  9. Question 9

    17 Apr 2026 S1

    72
    If y=[(x+1)(2x+1)(3x+1)……(nx+1)]4y = [(x + 1)(2x + 1)(3x + 1)\ldots\ldots(nx + 1)]^4, where n∈Nn \in N and dydx\dfrac{dy}{dx} at x=0x = 0 is 2k2k, then the value of kk is

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