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已知方程y=tan(y-x)确定了y=y(x) 求dy
人气:141 ℃ 时间:2020-09-14 05:25:59
解答
y=tan(y-x)
y'=[sec(y-x)]^2*(y'-1)
=[sec(y-x)]^2*y'-[sec(y-x)]^2
y'=[sec(y-x)]^2/ {[sec(y-x)]^2-1}
=1/[1-(cos(y-x))^2]
=[csc(y-x)]^2
dy=[csc(y-x)]^2dx
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