Chapter 1

\displaystyle \bold{(9.1)\quad} \lim_{x\to 0}\frac{\sin{x}}{x}=1\\

(COMING SOON…)

Chapter 2

\displaystyle \bold{(19.1)\quad} \text{Product Rule:\quad} \frac{d}{dx}(uv)=\frac{du}{dx}v+u\frac{dv}{dx}\\[2em]

\begin{aligned} \displaystyle(u(x)v(x))'&= \lim_{h\to 0}\frac{u(x+h)v(x+h)-u(x)v(x)}{h} \\[1em] &= \lim_{h\to 0}\frac{u(x+h)v(x+h)-u(x)v(x+h)+u(x)v(x+h)-u(x)v(x)}{h} \\[1em] &= \lim_{h\to 0}\left(\frac{u(x+h)-u(x)}{h}\cdot v(x+h)\right)+\lim_{h\to 0}\left( u(x)\cdot\frac{v(x+h)-v(x)}{h}\right)\\[1em] &=u'(x)v(x)+u(x)v'(x) \end{aligned} \\[2em]

\displaystyle \bold{(19.2)\quad} \text{Quotient Rule:\quad} \cfrac{d}{dx}(\cfrac{u}{v})=\cfrac{\cfrac{du}{dx}v-u\cfrac{dv}{dx}}{v^2}\\[2em]

\begin{aligned} \displaystyle \left(\frac{u(x)}{v(x)}\right)'&= \lim_{h\to 0}\cfrac{\cfrac{u(x+h)}{v(x+h)}-\cfrac{u(x)}{v(x)}}{h}\\[1em] &=\lim_{h\to 0}\cfrac{u(x+h)v(x)-u(x)v(x+h)}{v(x+h)v(x)}\cdot\frac{1}{h}\\[1em] &=\lim_{h\to 0}\cfrac{u(x+h)v(x)-u(x)v(x)+u(x)v(x)-u(x)v(x+h)}{h}\cdot\frac{1}{v(x+h)v(x)}\\[1em] &=\lim_{h\to 0}\left(\frac{u(x+h)-u(x)}{h}\cdot v(x)\cdot\frac{1}{v(x+h)v(x)}\right) \\[1em] &\quad+\lim_{h\to 0}\left( -u(x)\cdot\frac{v(x+h)-v(x)}{h}\cdot\frac{1}{v(x+h)v(x)}\right)\\[1em] &=u'(x)\cdot v(x)\cdot \frac{1}{v^2}-u(x)\cdot v'(x)\cdot \frac{1}{v^2}\\[1em] &=\frac{u'(x)v(x)-u(x)v'(x)}{v^2} \\[2em]\end{aligned}

\begin{aligned} \displaystyle \bold{(21.1)\quad} \text{If } y=f(g(x)) \text{, then } \frac{dy} {dx}=f'(g(x))\cdot g'(x) \end{aligned}

(COMING SOON…)

\displaystyle \bold{(22.1)\quad} \cfrac{d}{dx}(\sin{x})=\cos{x}\\[2em]

\begin{aligned} \displaystyle (\sin{x})'&= \lim_{h\to 0} \frac{\sin{(x+h)}-\sin{x}}{h} \\[1em] &=\lim_{h\to 0} \frac{\sin{x}\cos{h}+\cos{x}\sin{h}-\sin{x}}{h} \\[1em] &=\sin{x} \cdot \lim_{h\to 0} \frac{\cos{h}-1}{h} + \cos{x} \cdot \lim_{h\to 0} \frac{\sin{h}}{h} \end{aligned} \\[1em]

\text{In the first term,} \\[1em] \begin{aligned} \displaystyle \lim_{h\to 0} \frac{\cos{h}-1}{h}&= \lim_{h\to 0} \frac{(\cos{h}-1)(\cos{h}+1)}{h(\cos{h}+1)}\\[1em] &=\lim_{h\to 0} \frac{\cos^2{h}-1}{h(\cos{h}+1)}\\[1em] &=\lim_{h\to 0} \frac{-\sin^2{h}}{h(\cos{h}+1)}\\[1em] &=\lim_{h\to 0} (-\sin{h}) \cdot \underbrace{\lim_{h\to 0}\frac{\sin{h}}{h}}_\text{= 1 from (9.1)} \cdot \lim_{h\to 0}\frac{1}{\cos{h}+1}\\[1em] &=-0 \cdot 1 \cdot \frac{1}{2} =0 \end{aligned} \\[1em]

\text{Hence,} \\[1em] \begin{aligned} \displaystyle (\sin{x})'=\sin{x} \cdot 0 + \cos{x} \cdot 1 = \cos{x} \\[2em] \end{aligned}

\displaystyle \bold{(22.2)\quad} \cfrac{d}{dx}(\cos{x})=-\sin{x}\\[2em]

\begin{aligned} \displaystyle (\cos{x})'&=[ \underbrace{\sin{(\frac{\pi}{2}-x)}}_{\text{trig. identity}} ]' \\[2em] &= \cos{(\frac{\pi}{2}-x)} \cdot \underbrace{(\frac{\pi}{2}-x)'}_{\text{chain rule}} \\[1em] &= \sin{x} \cdot (-1) \\[1em] &=-\sin{x} \end{aligned} \\[2em]

\displaystyle \bold{(23.1)\quad} \cfrac{d}{dx}(e^x)=e^x\\[2em]

\displaystyle \text{The definition of } e \text{ is } e =\lim_{x\to \infty} ( 1+\frac{1}{x} )^x \\[2em] \text{Take natural logarithm for both side, and simplify}\\[1em] \begin{aligned} \ln{e}&=\lim_{x\to \infty} \ln{\left[ ( 1+\frac{1}{x} )^x\right]}\\[1em] 1&=\lim_{x\to \infty} x\ln{ ( 1+\frac{1}{x} )} \end{aligned} \\[2em] \text{let } u=\frac{1}{x}. \text{ When }x\to \infty, u\to 0 \text{. We have }\\[1em] \lim_{u\to 0} \frac{1}{u}\ln{ ( 1+u )} =1 \text{\qquad(*)} \\[2em] \text{Now we begin to find the derivative of } e^x \\[1em] \begin{aligned} (e^x)'&=\lim_{h\to 0}\frac{e^{x+h}-e^x}{h} \\[1em] &=e^x \cdot \lim_{h\to 0}\frac{(e^{h}-1)}{h} \end{aligned} \\[2em] \text{let } u=e^h-1 \text{, then } h=\ln{(u+1)} \text{. When }h\to 0, u\to 0 \text{. We have}\\[1em] \begin{aligned} (e^x)'&=e^x \cdot \lim_{h\to 0}\frac{(e^{h}-1)}{h}\\[1em] &= e^x \cdot \lim_{u\to 0} \frac{u}{\ln{(u+1)}}\\[1em] &= e^x \cdot \underbrace{\cfrac{1}{\displaystyle\lim_{u\to 0} \cfrac{\ln{(u+1)}}{u}}}_{\text{=1 from (*)}}\\[1em] &= e^x \end{aligned}