1个回答
2019-01-06
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题目似乎不全,这里只求导数。两边对x求导:
2x +2yy' = 2(2x² + 2y² - x)(2x² + 2y² - x)' = 2(2x² + 2y² - x)(4x + 4yy' - 1)
x + yy' = (2x² + 2y² - x)(4x - 1) + 4yy'(2x² + 2y² - x)
x - (2x² + 2y² - x)(4x - 1) = y'y[4(2x² + 2y² - x) - 1]
y' = [x - (2x² + 2y² - x)(4x - 1)]/[y(8x² + 8y² - 4x - 1)]
2x +2yy' = 2(2x² + 2y² - x)(2x² + 2y² - x)' = 2(2x² + 2y² - x)(4x + 4yy' - 1)
x + yy' = (2x² + 2y² - x)(4x - 1) + 4yy'(2x² + 2y² - x)
x - (2x² + 2y² - x)(4x - 1) = y'y[4(2x² + 2y² - x) - 1]
y' = [x - (2x² + 2y² - x)(4x - 1)]/[y(8x² + 8y² - 4x - 1)]
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