sin²x怎样求导 对SIN(x+y)求导
sin(x)\u7528\u6c42\u5bfc\u516c\u5f0f\u6c42\u5bfc\uff0c\u8981\u8be6\u7ec6\u6b65\u9aa4f\uff08x\uff09=sinx\uff0c\uff08f\uff08x+\u0394x\uff09-f\uff08x\uff09\uff09/\u0394x=\uff08sin\uff08x+\u0394x\uff09-sinx\uff09=2cos\uff08x+\u0394x/2\uff09sin\uff08\u0394x/2\uff09/\u0394x=
cos\uff08x+\u0394x/2\uff09sin\uff08\u0394x/2\uff09/\uff08\u0394x/2\uff09\u6240\u4ee5f'(x)=(sinx)'=lim(\u0394x\u21920\uff09\uff08f\uff08x+\u0394x\uff09-f\uff08x\uff09\uff09/\u0394x=cos\uff08x+\u0394x/2\uff09sin\uff08\u0394x/2\uff09/\uff08\u0394x/2\uff09=cosx \u5176\u4e2dlim(\u0394x\u21920\uff09sin\uff08(\u0394x/2\uff09/\uff08(\u0394x/2\uff09=1\u662f\u91cd\u8981\u6781\u9650\u3002
\u8fd9\u4e2a\u662f\u9690\u51fd\u6570\u7684\u6c42\u5bfc\u548c\u590d\u5408\u51fd\u6570\u7684\u6c42\u5bfc\u7684\u7ed3\u5408\u4e00\u5f00\u59cb\u4e0d\u662fcos(1+y')\u662f\u56e0\u4e3a\u524d\u9762\u662f\u4e00\u4e2a\u590d\u5408\u51fd\u6570\u6240\u4ee5\u5e94\u8be5\u662fcos(x+y)
f(x)=(sinx)^2令g(x)=sinx
f(g(x))=g(x)^2
f'(g(x))=2*g(x)
g'(x)=cosx
[f(g(x))]'=f'(g(x))*g'(x)=2*g(x)*g'(x)=2*sinx*cosx=sin2x
望采纳小女子答案为满意答案 谢谢
(sinx)^2=(1-cos2x)/2,故：(sinx)^2'=[(1-cos2x)/2]'=(2x)'(-cos2x)'/2=sin2x
(sinx^2)'=(cosx^2)*2x
((sinx)^2)'=2sinx*cosx
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