关于(sin(x))'=cos(x)的证明(sin(x))'=lim(h->0) (sin(x+h)-sin(x))/h,书上写的下一步是sin(x+h)-sin(x)=2cos(x+h/2)sin(h/2),然后再根据lim(h->0)sin(h/2)=lim(h->0)(h/2)得出(sin(x))'=cos(x)以下是我的疑问:{如果我第二步没有

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关于(sin(x))'=cos(x)的证明(sin(x))'=lim(h->0) (sin(x+h)-sin(x))/h,书上写的下一步是sin(x+h)-sin(x)=2cos(x+h/2)sin(h/2),然后再根据lim(h->0)sin(h/2)=lim(h->0)(h/2)得出(sin(x))'=cos(x)以下是我的疑问:{如果我第二步没有
关于(sin(x))'=cos(x)的证明
(sin(x))'=lim(h->0) (sin(x+h)-sin(x))/h,书上写的下一步是sin(x+h)-sin(x)=2cos(x+h/2)sin(h/2),然后再根据lim(h->0)sin(h/2)=lim(h->0)(h/2)得出(sin(x))'=cos(x)以下是我的疑问:{如果我第二步没有用和差化积,而是展开sin(x+h)那么得到lim(h->0) (sin(x)cos(h)+cos(x)sin(h)-sin(x))/h然后根据lim(h->0) cos(h)=1 得到上式=lim(h->0) cos(x)sin(h)/h再由lim(h->0)sin(h)=lim(h->0)(h)得到(sin(x))'=cos(x)☆我将cos(h)变为1,而没将sin(h)变成0是因为:cos(h)无论是1还是无限趋近于1的数字,除以分母的h,结果都是∞;而sin(h)变成0的话,0/h=0,非零的话,sin(h)/h=1,2种sin(h)的取值情况造成的结果不同,所以没有将其变为0.}以上是我的想法 我想知道是否可行?

关于(sin(x))'=cos(x)的证明(sin(x))'=lim(h->0) (sin(x+h)-sin(x))/h,书上写的下一步是sin(x+h)-sin(x)=2cos(x+h/2)sin(h/2),然后再根据lim(h->0)sin(h/2)=lim(h->0)(h/2)得出(sin(x))'=cos(x)以下是我的疑问:{如果我第二步没有
么么,这种看着头就晕