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ArCtAnx的导数

设x=tany tany'=sex^y arctanx'=1/(tany)'=1/sec^y sec^y=1+tan^y=1+x^2 所以(arctanx)'=1/(1+x^2)

解:y'=2arctanx*1/(1+x^2) =2arctanx/(1+x^2).

都换成反函数,再用复合函数求导法。 —————————————————————— y = arcsinx siny = x cosy * y' = 1 y' = 1/cosy = 1/√(1 - sin²y) = 1/√(1 - x²) —————————————————————— y = arccosx cosy = x - siny * y' = 1 y' = - 1/siny = - 1/√...

1/(1+x^2)

1)y1 = arctanx..........y'1 = 1/(1+x²) 2) y2 = arccotx...........y'2 = -1/(1+x²) 3) 可见:y'1 = - y'2 4) y = arctanx tany = x y' sec²y = 1 y' = 1/sec²y = 1/(1+x²)..........1+tan²y = sec²y = 1...

[-arctanx]′ = -1/(1+x²)

y=arctanx,则x=tany arctanx′=1/tany′ tany′=(siny/cosy)′=cosycosy-siny(-siny)/cos²y=1/cos²y 则arctanx′=cos²y=cos²y/sin²y+cos²y=1/1+tan²y=1/1+x² 故最终答案是1/1+x² 希望能帮到你

y=arctanx 那么tany=x 求导得到 1/cos²y*y'=1 即y'=cos²y=1/(1+x²)

在详细的我也说不出来了 下面的写的参考看看 y=arctanx,则x=tany arctanx′=1/tany′ tany′=(siny/cosy)′=cosycosy-siny(-siny)/cos²y=1/cos²y 则arctanx′=cos²y=cos²y/sin²y+cos²y=1/1+tan²y=1/1+x²

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