求sinx的n阶导数，麻烦给个过程 求y=(sinx)^的n阶导数? (带上过程)
\u6c42sinx\u7684n\u9636\u5bfc\u6570,\u9ebb\u70e6\u7ed9\u4e2a\u8fc7\u7a0b(sinx)'=cosx=sin(x+\u03c0/2)
(sinx)''=[sin(x+\u03c0/2)]'=cos[x+(\u03c0/2)]=sin[x+2(\u03c0/2)]
\u2026\u2026
\uff08sinx)^(n)=[sin(x+(n-1)(\u03c0/2))]'=cos[x+(n-1)(\u03c0/2)]=sin[x+n(\u03c0/2)]
y=sin²x=(1/2)(1-cos2x)
y'=(1/2)*2sin(2x)=sin(2x)
y''=2cos(2x)=2sin(2x+\u03c0/2)
y'''=-4sin(2x)=4sin(2x+\u03c0)
y^(4)=-8cos(2x)=8sin(2x+3\u03c0/2)
y^(5)=16sin(2x)=16sin(2x+2\u03c0)
.......
y^(n)=[2^(n-1)]sin(2x+(n-1)\u03c0/2)
\u5e0c\u671b\u53ef\u4ee5\u5e2e\u5230\u4f60\uff0c\u4e0d\u660e\u767d\u53ef\u4ee5\u8ffd\u95ee\uff0c\u5982\u679c\u89e3\u51b3\u4e86\u95ee\u9898\uff0c\u8bf7\u70b9\u4e0b\u9762\u7684"\u9009\u4e3a\u6ee1\u610f\u56de\u7b54"\u6309\u94ae\u3002
(sinx)''=[sin(x+π/2)]'=cos[x+(π/2)]=sin[x+2(π/2)]
……
（sinx)^(n)=[sin(x+(n-1)(π/2))]'=cos[x+(n-1)(π/2)]=sin[x+n(π/2)]
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