y' = 1 - 2x/(5 + x²) = (x² - 2x + 5)/(x² + 5)
x² + 5 > 0
x² - 2x + 5 = (x - 1)² + 4 > 0
函数在定义域(-∞,+∞)内递增
你把开始的x漏了.
函数y=x-ln(5+x^2)的单调增区间为_____.
0x² - 2x + 5 = (x - 1"}}}'>
y' = 1 - 2x/(5 + x²) = (x² - 2x + 5)/(x² + 5)
x² + 5 > 0
x² - 2x + 5 = (x - 1)² + 4 > 0
函数在定义域(-∞,+∞)内递增
你把开始的x漏了.