prove cosx is the derivative of sinx

我们一直都在使用Derivative Rules，但有没有想过为什么是这样的？

方1

由导数定义有
$\frac{d}{dx}sin(x)=\underset{h\to 0}{\lim }\frac{sin(x+h)-sin(x)}{h}$
$\frac{d}{dx}sin(x)=\underset{h\to 0}{\lim }\frac{sin(x)cos(h)+cos(x)sin(h)-sin(x)}{h}$
$\frac{d}{dx}sin(x)=cos(x)\underset{h\to 0}{\lim }\frac{sin(h)}{h}-sin(x)\underset{h\to 0}{\lim }\frac{1-cos(h)}{h}$
证明$(sinx{)}^{\prime }=cosx$就相当于证明$\underset{h\to 0}{\lim }\frac{sin(h)}{h}=1,\underset{h\to 0}{\lim }\frac{1-cos(h)}{h}=0$

$\frac{1-cos(x)}{x}=\frac{sinx}{x}\frac{sinx}{1+cosx}$
$\underset{h\to 0}{\lim }\frac{sinx}{x}\frac{sinx}{1+cosx}=0$

参考
https://www.analyzemath.com/calculus/derivative/proof-derivative-of-e%5Ex.html

方2

由欧拉公式有
$sinx=\frac{1}{2i}(e^{ix}-e^{-ix})$
$cosx=\frac{1}{2}(e^{ix}+e^{-ix})$
证明$(sinx{)}^{\prime }=cosx$就相当于证明$(e^{cx}{)}^{\prime }=c{e}^{cx}$

$(e^{cx}{)}^{\prime }=\underset{h\to 0}{\lim }\frac{e^{c(x+h)}-e^{cx}}{h}$
$(e^{cx}{)}^{\prime }=c{e}^{cx}\underset{h\to 0}{\lim }\frac{e^{ch}-1}{ch}$
$\underset{h\to 0}{\lim }\frac{e^{ch}-1}{ch}=\underset{y\to 0}{\lim }\frac{y}{ln(y+1)},y=e^{ch}-1$
$\underset{y\to 0}{\lim }\frac{y}{ln(y+1)}=\underset{y\to 0}{\lim }ln(y+1{)}^{-\frac{1}{y}}$
欧拉常数 $e=\underset{m\to 0}{\lim }(1+m{)}^{\frac{1}{m}},ln(e)=1$
$\underset{y\to 0}{\lim }ln(y+1{)}^{-\frac{1}{y}}=\frac{1}{ln(\underset{y\to 0}{\lim }(y+1{)}^{\frac{1}{y}})}=1$
因此，$(e^{cx}{)}^{\prime }=c{e}^{cx}$

参考
https://www.analyzemath.com/calculus/derivative/proof-derivative-of-e%5Ex.html

还可以通过级数推导

Note:
the logarithm is the inverse function to exponentiation
if $lo{g}_{a}b=c$, then $a^c=b$