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E的%(ArCtAnx)²的导数怎么算呀?

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解:令y=arctanx,则x=tany。 对x=tany这个方程“=”的两边同时对x求导,则 (x

y=arctanx,则x=tany arctanx′=1/tany′ tany′=(siny/c

(e^arctanx)'=(e^arctanx)*(arctanx)'

x=tany y= arctanx dx/dy =1/sec^2(y)=1/(1+tan^2(y

解: y=arctanx,则x=tany arctanx′=1/tany′ tany′=(siny/

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

y=arctanx,则x=tany arctanx′=1/tany′ tany′=(siny&a

(2arctanx+arcsinx)' =2/(1+x²)+√(1+x&

[-arctanx]′ = -1/(1+x²)

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