If xy + 4ey = 4e, find the value of y” at the point where x = 0.?

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If xy + 4ey = 4e, find the value of y” at the point where x = 0.

cannot figure out why I keep getting it wrong.

please show steps if possible

2 Answers

  • xy + 4ey = 4e

    y(x + 4e) = 4e

    y = 4e / (x + 4e)

    y’ = -4e / ((x + 4e)^2)

    y” = 8e(x + 4e) / ((x + 4e)^4)

    So when x = 0, y” = (32e^2)/((4e)^4) = (32e^2)/(16e^4) = 2/(e^2).

  • Answer is 1/(16e^2)

    If you have the xy+2e^y=2e variant, then it’s 1/(4e^2). If you have the xy+3e^y=3e variant, then it’s 1/(9e^2), and so on and so forth.

    Source(s): I got it wrong on WebAssign, so they showed me the correct answer. I don’t actually know the correct method to solve this. but I thought I would help out all those who don’t have access to Chegg.

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