求y=arcsin(sinx)^(1/2)

题目

求y=arcsin(sinx)^(1/2)
我算得1/2*{1/[(1-sinx)*sinx]}^(1/2),答案是1/2*(1+cscx)^(1/2)

答案

(sinx)'=cosx
[(sinx)^(1/2)]'=(1/2)(sinx)^(-1/2)
[arcsin(sinx)^(1/2)]'=1/(1-sinx)^(1/2)
y'=(1/2)cosx*(sinx)^(-1/2)*[1/(1-sinx)^(1/2)
=(1/2)cosx*(sinx)^(-1/2)*[(1+sinx)^(1/2)/(1-(sinx)^2)^(1/2)]
=(1/2)cosx*(sinx)^(-1/2)*[(1+sinx)^(1/2)/cosx]
=(1/2)(sinx)^(-1/2)*(1+sinx)^(1/2)
=(1/2)[(1+sinx)/sinx]^(1/2)
=(1/2)(1/sinx+1)^(1/2)
=(1/2)(cscx+1)^(1/2)