sin2x的求导是什么？就是用只有x的
\u6c42\u52a9 \uff08sin2x\uff09/x \u6c42\u5bfc \u8c22\u8c22\u4e86[\uff08sin2x\uff09/x ]'
=(2xcos2x-sin2x)/x^2
(\u8fd0\u7528\u516c\u5f0f(u/v)'=(u'v-uv')/v^2)
f'(x)=(2xcos2x-sin2x)/x^2
\u5229\u7528\u9664\u6cd5\u6c42\u5bfc\u6cd5\u5219 (u/v)'=(u'v-uv')/v^2
\u8fc7\u7a0bf'(x)=[(sin2x)'x - x'sin2x]/x^2=(2xcos2x-sin2x)/x^2
使用基本求导公式，一步步来求导即可
那么就得到
(sin2x)'
=cos2x *(2x)'
=cos2x *2
=2cos2x
2cos2x
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