y=反正切x+1/x-1的导数
2个回答
2016-10-31
展开全部
y=arctan[(x+1)/(x-1)]
y′ = 1/{1+[(x+1)/(x-1)]² } * [(x+1)/(x-1)]′
= 1/{1+[(x+1)/(x-1)]² } * [1 + 2/(x-1)]′
= 1/{1+[(x+1)/(x-1)]² } * [2/(x-1)]′
= 1/{1+[(x+1)/(x-1)]² } * {-2/(x-1)²}
= -2 / {(x-1)²+[(x+1)² }
= - 1 / (x²+1)
y′ = 1/{1+[(x+1)/(x-1)]² } * [(x+1)/(x-1)]′
= 1/{1+[(x+1)/(x-1)]² } * [1 + 2/(x-1)]′
= 1/{1+[(x+1)/(x-1)]² } * [2/(x-1)]′
= 1/{1+[(x+1)/(x-1)]² } * {-2/(x-1)²}
= -2 / {(x-1)²+[(x+1)² }
= - 1 / (x²+1)
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