高数,求详解,第三题第一个
1个回答
2016-06-29
展开全部
Zx=[-yf'(x²-y²)•2x]/f²(x²-y²)
=[-2xyf'(x²-y²)]/f²(x²-y²)
Zy=[f(x²-y²) -yf'(x²-y²)•(-2y)]/f²(x²-y²)
=1/f²(x²-y²) +2y²f'(x²-y²)/f²(x²-y²)
=[-2xyf'(x²-y²)]/f²(x²-y²)
Zy=[f(x²-y²) -yf'(x²-y²)•(-2y)]/f²(x²-y²)
=1/f²(x²-y²) +2y²f'(x²-y²)/f²(x²-y²)
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