sin(x)导数求解
导数定义 $f’(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$ 对于sin(x)的导数 $\frac{d}{dx} \sin(x) = \lim_{h \to 0} \frac{\sin(x+h) - \sin(x)}{h}$ 由三角函数的和角公式 $\sin(x + h) = \sin(x) \cos(h) + \cos(x) \sin(h)$ 将其带入导数的定义中: $\frac{\sin(x+h) - \sin(x)}{h} = \frac{\sin(x) \cos(h) + \cos(x) \sin(h) - \sin(x)}{h}$ 可以分解为: $\frac{\sin(x+h) - \sin(x)}{h} = \frac{\sin(x) \cos(h) - \sin(x) + \cos(x) \sin(h)}{h}= \frac{\sin(x) (\cos(h) - 1)}{h} + \cos(x) \frac{\sin(h)}{h}$ 极限的计算 我们现在需要计算两个极限: 第一个极限:$(\lim_{h \to 0} \frac{\sin(x) (\cos(h) - 1)}{h})$ 当$ (h \to 0)$ 时,$(\cos(h) \to 1)$,因此 $(\cos(h) - 1 \to 0)$,并且 $(\frac{\cos(h) - 1}{h})$ 的极限是 0。具体地: ...