arctanx'=1/(1+x^2)
y=arctan(x+1)^1/2
y'=1/(1+(x+1)^1/2^2)*(x+1)^1/2'
y'=1/(x+2)*1/2(x+1)^(-1/2)
y'=1/[2(x+2)根号(x+1)]
y=arctan(x+1)^1/2,求dy=?
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