第3课：
- 导数的加法法则：$(u + v)’ = u’ + v’$
- 导数的数乘法则：$(cu)’ = cu’$
- 可去间断点：$$A: \;\;\;\;g(x) = \frac{sin(x)}{(x)},\;\;\;\;\lim_{x\rightarrow 0}\frac{sin(x)}{x}=1$$
$$B:\;\;\;\; h(x) = \frac{1-cos(x)}{x},\;\;\;\;\;\lim_{x\rightarrow 0}\frac{1-cos(x)}{x} = 0$$ - $sin(x)’ = cos(x)$和$(cosx)’= -sin(x)$的证明：$$\frac{sin(x+\Delta x) - sin(x)}{\Delta x} = \frac{sinxcos\Delta x + cosxsin\Delta x - sinx}{\Delta x}\=sinx(\frac{cos \Delta x - 1}{\Delta x}) + cosx(\frac{sin \Delta x}{\Delta x})\=cosx \ ————————————–\ \frac{cos(x + \Delta x)- cos x}{\Delta x}=\frac{cosxcos\Delta x - sinx sin \Delta x - cosx}{\Delta x}\=cosx(\frac{cos \Delta x - 1}{\Delta x})- sinx(\frac{sin \Delta x}{\Delta x})\=-sinx$$
使用极限几何证明A、B。
证明 A:
$$\Delta x \rightarrow \Theta,\;\;\;\;\;\;\;\; \frac{弓弦}{弓} = \frac{2 sin \Theta}{2 \Theta} \rightarrow 1$$注：极短的曲线可以看作直线。
- 证明B:
$$\frac{1-cos\Theta}{\Theta}\rightarrow 0,\;\;\;\;\;\;\; 前提：这个圆还是单位圆。当红色部分很小时，即\Theta 很小的时候，等式便为0。$$
- 证明：$\frac{\Delta y}{ \Delta \Theta} = cos\Theta$
$$OR \perp PQ ,QR \perp PR ,\Delta y = PR, PQ \approx \Delta \Theta,\frac{\Delta y}{ \Delta \Theta} = cos\Theta$$
- $(\frac{u}{v})’ = \frac{(u’v - uv’)}{v^2}\;\;\;\;\;\;\;\;(v \neq 0)$
- $(uv)’=(u’v+uv’)$
$Pf:$
$$\Delta (uv) = u(x + \Delta x)v(x+\Delta x) - u(x)v(x)\=(u(x+\Delta x) - u(x))v(x+\Delta x)+u(x)(v(x+\Delta x)- v(x))\=\Delta uv(x+\Delta x)+u(x)\Delta v \ \frac {\Delta (uv)}{\Delta x}= \frac{\Delta u}{\Delta x}v(x + \Delta x) + \frac{u \Delta v}{ \Delta x}\由于\Delta x \rightarrow 0 所以，\ \frac{d(uv)}{dx}=\frac{du}{dx} \cdot v+ u \frac{dv}{dx}$$
$$\Delta (\frac{u}{x})=\frac{u + \Delta x}{v + \Delta v}-\frac{u}{v}\=\frac{uv+(\Delta u)v- uv - u\Delta v}{(v + \Delta v)v}\=\frac{(\Delta u)v- u \Delta v}{(v + \Delta v)v}\\frac{\Delta(\frac{u}{x})}{\Delta x}=\frac{\frac{\Delta u}{\Delta x}v-u\frac{\Delta v}{\Delta x}}{(v + \Delta v )v}\由于\Delta x \rightarrow 0,所以\\frac{d}{dx}(\frac{u}{v})=\frac{\frac{du}{dx} \cdot v - u \frac{dv}{dx}}{v \cdot v}$$