sin(x)的导数推导问题
\u5e2e\u5fd9\u63a8\u5bfc\u51fd\u6570f(x)=sin x\u7684\u5bfc\u6570\u516c\u5f0f\u89e3\u7b54\uff1a
f(x+a)-f(x)=sin(x+a)-sinx=2cos(x+0.5a)sin0.5a
f(x+a)-f(x)/[(x+a)-x]=cos(x+0.5a)sin0.5a/0.5a
\u663e\u7136\u67090<sinx<x<tanx x\u4e3a\u9510\u89d2 \u901a\u8fc7\u6781\u9650\u5939\u6324\u51c6\u5219\u6613\u5f97limx/sinx=limsinx/x=1
\u6240\u4ee5f'(x)=(sinx)'=
limf(x+a)-f(x)/[(x+a)-x]=
limcos(x+0.5a)sin0.5a/0.5a=limcos(x+0.5a)*limsin0.5a/0.5a
=cosx*1=cosx
\u6240\u4ee5\u82e5f(x)=sin x\u3001
\u5219\u6709f\u2018(x)=cosx
\u7b2c\u4e00\u9898\uff1a\u6309\u7167\u4e09\u89d2\u51fd\u6570\u516c\u5f0f\u548c\u5bfc\u6570\u7684\u5b9a\u4e49\u5c31\u53ef\u4ee5\u8bc1\u660e
lim(\u0394y/\u0394x)
\u0394x->0
=lim{[sin(x+\u0394x)-sin(x)]/\u0394x}
\u0394x->0
=lim[2cos(x+\u0394x/2)sin(\u0394x/2)/\u0394x]
\u0394x->0
=lim[cos(x+\u0394x/2)sin(\u0394x/2)/\u0394x/2]
\u0394x->0
\u7531cos(x)\u7684\u8fde\u7eed\u6027\uff0c\u6709limcos(x+\u0394x/2) = cos(x)
\u0394x->0
\u4ee5\u53calim[sin(\u0394x/2)/\u0394x/2] = 1
\u0394x->0
\u6545\u5f97
lim(\u0394y/\u0394x)
\u0394x->0
=limcos(x+\u0394x/2)*lim[sin(\u0394x/2)/\u0394x/2]
\u0394x->0 \u0394x->0
=cos(x)*1
=cos(x)
\u7b2c\u4e8c\u9898\uff1a
\uff081\uff09\u5229\u7528\u5bfc\u6570\u7684\u5b9a\u4e49\uff1a
[cos(x)]'=lim{[cos(x+h)-cos(x)]/h}
\u6ce8\u610f\uff1a\u6781\u9650\u8fc7\u7a0b\u662fh\u21920
\uff082\uff09\u5229\u7528\u4e09\u89d2\u516c\u5f0f\u4e2d\u7684\u548c\u5dee\u5316\u79ef\u516c\u5f0f\uff1a
[cos(x)]'=lim{[cos(x+h)-cos(x)]/h}
=lim{(1/h)*[-2sin(x+h/2)*sin(h/2)]}
=lim{-sin(x+h/2)*[sin(h/2)/(h/2)]}
\uff083\uff09\u5728\u9ad8\u6570\u6781\u9650\u4e00\u7ae0\u6211\u4eec\u5df2\u7ecf\u719f\u77e5\u7684\u91cd\u8981\u6781\u9650:
lim[sin(x)/x]=1(\u6781\u9650\u8fc7\u7a0b\u662fx\u21920)
\uff084\uff09[cos(x)]'=-sin(x),\u5f97\u8bc1\u3002
\u5982\u679c\u6ee1\u610f\u8bb0\u5f97\u91c7\u7eb3\u54e6\uff01
\u4f60\u7684\u597d\u8bc4\u662f\u6211\u524d\u8fdb\u7684\u52a8\u529b\u3002
(*^__^*) \u563b\u563b\u2026\u2026
\u6211\u5728\u6c99\u6f20\u4e2d\u559d\u7740\u53ef\u53e3\u53ef\u4e50\uff0c\u5531\u7740\u5361\u62c9ok\uff0c\u9a91\u7740\u72ee\u5b50\u8d76\u7740\u8682\u8681\uff0c\u624b\u4e2d\u62ff\u7740\u952e\u76d8\u4e3a\u4f60\u7b54\u9898\uff01\uff01\uff01
我是数学系大三学生，祝你好运！Good luck with you!
d->0就是意味着将d=0代入式中
就是cos[x+(0/2)]*sin(0/2)(0/2)=cosx*sin0=cosx
sin0=1这个你应该知道吧...
初等数学中有这样一个不等式
如果0<x<π/2 则有 sinx<x<tanx （这个不等式可以通过画出三角函数线 简单计算得出）
使用上式 有1<x/sinx<1/cosx
当x趋近于0的时候1/cosx趋近于1
通过极限的夹挤准则可知 当x趋近于0的时候lim x/sinx=lim sinx/x=1
当dx->0时，dx/2->0,所以sin(dx/2)/(dx/2)->1
当然 高等数学中把这个极限作为了2个重要极限，但是推导的主要原理就是这样了~~
祝学习进步~~
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