differentiation

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find dy/dx if xy log(x+y)=1

  • differentiation -

    looks like triple product rule
    recall
    d(uvw) = uvw' + uwv' + vwu'

    I will also assume that you meant ln(x+y)

    xy ln(x+y) = 1
    xy(1/(x+y)) * (1 + dy/dx) + x ln(x+y) dy/dx + y ln(x+y) = 0

    expand the first part to get at the dy/dx
    xy(1/(x+y)) + dy/dx xy(1/(x+y)) + dy/dx (x ln(x+y)) = -y ln(x+y) - xy(1/(x+y))
    dy/dx [xy(1/(x+y)) + (x ln(x+y)) ] = -y ln(x+y) - xy(1/(x+y))
    dy/dx = (-y ln(x+y) - xy(1/(x+y))) / [xy(1/(x+y)) + (x ln(x+y)) ]

    leave it like that or simplify as needed.

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