设z=yf(x2-y2),其中f(u)可微,则=_______.

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问题 设z=yf(x2-y2),其中f(u)可微,则=_______.

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答案z/y2

解析 =yf’(x2-y2).2x=2xyf’(x2-y2),
=yf’(x2-y2)(-2y)+f(x2-y2)=-2y2f’(x2-y2)+f(x2-y2),
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