求由方程xy=e^x+y所确定的隐函数y=y(x)的导数

Question

如题所述

Answer 1

xy=e^(x+y)
两边求导：
y + xy ′ = e^(x+y) * (1+y ′)
y + xy ′ = e^(x+y) + e^(x+y) * y ′
xy ′ - e^(x+y) * y ′ = e^(x+y) - y
y ′ = {e^(x+y) - y} / { x - e^(x+y) }

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xy=e^x+y
两边求导：
y + xy ′ = e^x + y ′
xy ′ - y ′ = e^x - y
y ′ = ( e^x - y ) / (x-1)