求y=arcsin(sinx)^(1/2)

学习 时间:2026-08-11 09:44:01 阅读:3055
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)

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强健的小兔子

苗条的航空

2026-08-11 09:44:01

(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)

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  • 尊敬的乌龟
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    2026-08-11 09:44:01

    (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)

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