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2 February, 16:49

If y=u + 2e^u and u=1+lynx

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Answers (2)
  1. 2 February, 17:45
    0
    I am not very good at math but i wanna say this is how you do it not a 100% sure

    y = u + 2e^u

    dy/du = 1 + 2e^u

    u = 1 + ln x

    du/dx = 1/x

    Then dy/dx = dy/du * du/dx

    dy/dx = (1+2e^u) * 1/x

    Now replace u with 1 + ln x

    dy/dx = (1 + 2e^ (1+lnx)) * 1/x

    dy/dx = (1 + 2e^1 * e^ln x) * 1/x

    dy/dx = (1 + 2e x) * 1/x

    dy/dx = 1/x + 2e

    Now let x = 1/e. Then

    dy/dx = 1 / (1/e) + 2e = e + 2e = 3e
  2. 2 February, 18:48
    0
    Y = u + 2e^u

    dy/du = 1 + 2e^u

    u = 1 + ln x

    du/dx = 1/x

    Then dy/dx = dy/du * du/dx

    dy/dx = (1+2e^u) * 1/x

    Now replace u with 1 + ln x

    dy/dx = (1 + 2e^ (1+lnx)) * 1/x

    dy/dx = (1 + 2e^1 * e^ln x) * 1/x

    dy/dx = (1 + 2e x) * 1/x

    dy/dx = 1/x + 2e

    Now let x = 1/e. Then

    dy/dx = 1 / (1/e) + 2e = e + 2e = 3e
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