有关y={(x-4)(x-3)/[(x-2)(x-1)]}^1/2求导数的问题.书上两边取对数,得到lny=1/2[ln(x-4)+ln(x-3)-ln(x-2)-ln(x-1)],这里为什么不考虑x<1的情况?
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![有关y={(x-4)(x-3)/[(x-2)(x-1)]}^1/2求导数的问题.书上两边取对数,得到lny=1/2[ln(x-4)+ln(x-3)-ln(x-2)-ln(x-1)],这里为什么不考虑x<1的情况?](/uploads/image/z/4296866-50-6.jpg?t=%E6%9C%89%E5%85%B3y%3D%7B%28x-4%29%28x-3%29%2F%5B%28x-2%29%28x-1%29%5D%7D%5E1%2F2%E6%B1%82%E5%AF%BC%E6%95%B0%E7%9A%84%E9%97%AE%E9%A2%98.%E4%B9%A6%E4%B8%8A%E4%B8%A4%E8%BE%B9%E5%8F%96%E5%AF%B9%E6%95%B0%2C%E5%BE%97%E5%88%B0lny%3D1%2F2%5Bln%28x-4%29%2Bln%28x-3%29-ln%28x-2%29-ln%28x-1%29%5D%2C%E8%BF%99%E9%87%8C%E4%B8%BA%E4%BB%80%E4%B9%88%E4%B8%8D%E8%80%83%E8%99%91x%EF%BC%9C1%E7%9A%84%E6%83%85%E5%86%B5%3F)
有关y={(x-4)(x-3)/[(x-2)(x-1)]}^1/2求导数的问题.书上两边取对数,得到lny=1/2[ln(x-4)+ln(x-3)-ln(x-2)-ln(x-1)],这里为什么不考虑x<1的情况?
有关y={(x-4)(x-3)/[(x-2)(x-1)]}^1/2求导数的问题.
书上两边取对数,得到lny=1/2[ln(x-4)+ln(x-3)-ln(x-2)-ln(x-1)],这里为什么不考虑x<1的情况?
有关y={(x-4)(x-3)/[(x-2)(x-1)]}^1/2求导数的问题.书上两边取对数,得到lny=1/2[ln(x-4)+ln(x-3)-ln(x-2)-ln(x-1)],这里为什么不考虑x<1的情况?
y={(x-4)(x-3)/[(x-2)(x-1)]}^1/2的定义域是(-∞,1)∪(2,3]∪[4,+∞)
x≥4时,lny=1/2[ln(x-4)+ln(x-3)-ln(x-2)-ln(x-1)],
y′/y=1/2[1/(x-4)+1/(x-3)-1/(x-2)-1/(x-1)].(*)
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