由隐函数微分法可得：－sin(x＋y)(1＋y′ )＋y′ ＝0
－sin(x＋y)＋[1－sin(x＋y)]y′ ＝0
∴y′ ＝sin(x＋y)／[1－sin(x＋y)]．