導數

• 瞬時變化率，有「速度」為證。夫所謂以 ${\displaystyle {\vec {v}}}$  之速而行者，實是頃間位移之變，亦位移之導數而已矣。故曰：${\displaystyle {\frac {\mathrm {d} {\vec {s}}}{\mathrm {d} t}}={\vec {v}}}$ 。

設 ${\displaystyle I}$  為一開區間且函數 ${\displaystyle f:I\to \mathbb {R} }$ ，${\displaystyle x\in I}$ ，若極限

　　${\displaystyle \lim _{\Delta x\to 0}{\frac {f(x+\Delta x)-f(x)}{\Delta x}}}$ 

存在，是謂 ${\displaystyle f}$  可微分於點 ${\displaystyle x}$  。其極限值，即 ${\displaystyle f}$  微分值之在 ${\displaystyle x}$  ，且如上述云。

設 ${\displaystyle I}$  為一開區間且 ${\displaystyle f:I\to \mathbb {R} }$  上處處可微分於 ${\displaystyle I}$  ，則命 ${\displaystyle f}$  之導數 ${\displaystyle f'(x)}$  於 ${\displaystyle I}$  ：

${\displaystyle f'(x):=\lim _{\Delta x\to 0}{\frac {f(x+\Delta x)-f(x)}{\Delta x}}}$ 　　　　${\displaystyle \forall x\in I}$ 

亦以 ${\displaystyle {\frac {\mathrm {d} f(x)}{\mathrm {d} x}}}$  識之。

顯函之導數亦 ${\displaystyle x}$  之一函數也，隱函者則為 ${\displaystyle \mathbb {R} ^{n}}$  空间之一多元函數也。

凡以下公式，皆助吾等得眾函之導數：

1.${\displaystyle y=c,y'=0}$ 

2.${\displaystyle y=x,y'=1}$ 

3.${\displaystyle y=x^{n},y'=nx^{n-1}}$ 

4.${\displaystyle y=e^{x},y'=e^{x}}$ 

5.${\displaystyle y=\ln |x|,y'={\frac {1}{x}}}$ 

6.${\displaystyle y=\sin(x),y'=\cos(x)}$ 

7.${\displaystyle y=\cos(x),y'=-\sin(x)}$ 

8.${\displaystyle y=\tan(x),y'=\sec ^{2}(x)}$ 

9.${\displaystyle y=\csc(x),y'=-\csc(x)\cot(x)}$ 

10.${\displaystyle y=\sec(x),y'=\sec(x)\tan(x)}$ 

11.${\displaystyle y=\cot(x),y'=-\csc ^{2}(x)}$ 

12.${\displaystyle y=\arcsin(x),y'={\frac {1}{\sqrt {1-x^{2}}}}}$ 

13.${\displaystyle y=\arccos(x),y'=-{\frac {1}{\sqrt {1-x^{2}}}}}$ 

14.${\displaystyle y=\arctan(x),y'={\frac {1}{1+x^{2}}}}$ 

15.${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}(u\pm v)={\frac {\mathrm {d} u}{\mathrm {d} x}}\pm {\frac {\mathrm {d} v}{\mathrm {d} x}}}$  （加減之法）

16.${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}(uv)=u{\frac {\mathrm {d} v}{\mathrm {d} x}}+v{\frac {\mathrm {d} u}{\mathrm {d} x}}}$  （乘之法）

17.${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}{\frac {u}{v}}={\frac {v{\frac {\mathrm {d} u}{\mathrm {d} x}}-u{\frac {\mathrm {d} v}{\mathrm {d} x}}}{v^{2}}}}$  （除之法）

18.${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}y(u(x))={\frac {\mathrm {d} y}{\mathrm {d} u}}{\frac {\mathrm {d} u}{\mathrm {d} x}}}$  （鏈之法）

夫初等函數之萬千组合，毋論顯隱，此眾法皆可得其導數。

證

一之證： ${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}(C)}$ 

${\displaystyle =\lim _{\Delta x\to 0}{\frac {C-C}{\Delta x}}}$ 

${\displaystyle =0}$ 

二之證： ${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}(x)}$ 

${\displaystyle =\lim _{\Delta x\to 0}{\frac {(x+\Delta x)-x}{\Delta x}}}$ 

${\displaystyle =\lim _{\Delta x\to 0}{\frac {\Delta x}{\Delta x}}}$ 

${\displaystyle =1}$ 

三之證：

• 次第一：${\displaystyle n}$  乃正整數，藉乎牛頓二項式定理：

${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}(x^{n})}$ 

${\displaystyle =\lim _{\Delta x\to 0}{\frac {(x+\Delta x)^{n}-x^{n}}{\Delta x}}}$ 

${\displaystyle =\lim _{\Delta x\to 0}{\frac {x^{n}+C_{1}^{n}x^{n-1}\Delta x+C_{2}^{n}x^{n-2}\Delta x^{2}+...+C_{n}^{n}\Delta x^{n}-x^{n}}{\Delta x}}}$ 

${\displaystyle =\lim _{\Delta x\to 0}C_{1}^{n}x^{n-1}+C_{2}^{n}x^{n-2}\Delta x+...+C_{n}^{n}\Delta x^{n-1}}$ 

${\displaystyle =C_{1}^{n}x^{n-1}}$ 

${\displaystyle =nx^{n-1}}$ 

• 次第二：${\displaystyle n}$  乃負整數

設 ${\displaystyle k=-n}$  。則： ${\displaystyle x^{n}={\frac {1}{x^{k}}}}$ 

${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}(x^{n})}$ 

${\displaystyle ={\frac {\mathrm {d} }{\mathrm {d} x}}{\frac {1}{x^{k}}}}$ 

${\displaystyle ={\frac {x^{k}{\frac {\mathrm {d} }{\mathrm {d} x}}(1)-1{\frac {\mathrm {d} }{\mathrm {d} x}}(x^{k})}{x^{2k}}}}$ 

${\displaystyle ={\frac {0-kx^{k-1}}{x^{2k}}}}$ 

${\displaystyle =-kx^{-k-1}}$ 

${\displaystyle =nx^{n-1}}$ 

• 次第三：${\displaystyle n}$  乃有理數

設 ${\displaystyle n={\frac {p}{q}}}$  ，且 ${\displaystyle p,q}$  乃整數 （${\displaystyle q\neq 0}$ ）。則： ${\displaystyle x^{n}=(x^{\frac {1}{q}})^{p}}$ 

又：設 ${\displaystyle u=x^{\frac {1}{q}}}$  ，則： ${\displaystyle y=u^{p}}$ 

${\displaystyle {\frac {\mathrm {d} y}{\mathrm {d} u}}=pu^{p-1}}$ 

${\displaystyle {\frac {\mathrm {d} x}{\mathrm {d} u}}=qu^{q-1}}$ 

${\displaystyle {\frac {\mathrm {d} y}{\mathrm {d} u}}={\frac {\mathrm {d} y}{\mathrm {d} x}}{\frac {\mathrm {d} x}{\mathrm {d} u}}}$ 

${\displaystyle pu^{p-1}={\frac {\mathrm {d} }{\mathrm {d} x}}(x^{n}).qu^{q-1}}$ 

${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}(x^{n})={\frac {p}{q}}u^{p-q}}$ 

${\displaystyle ={\frac {p}{q}}(x^{\frac {1}{q}})^{p-q}}$ 

${\displaystyle =nx^{n-1}}$ 

• 次第四： ${\displaystyle n}$  乃實數

設 ${\displaystyle y=x^{n}}$  。則： ${\displaystyle \ln |y|=n\ln |x|}$ 

${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}(\ln |y|)={\frac {\mathrm {d} }{\mathrm {d} x}}n\ln |x|}$ 

${\displaystyle {\frac {1}{y}}.{\frac {\mathrm {d} y}{\mathrm {d} x}}=n.{\frac {1}{x}}}$ 

${\displaystyle {\frac {\mathrm {d} y}{\mathrm {d} x}}=y.{\frac {n}{x}}}$ 

${\displaystyle =x^{n}.{\frac {n}{x}}}$ 

${\displaystyle =nx^{n-1}}$ 

四之證：

${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}(e^{x})}$ 

${\displaystyle =\lim _{\Delta x\to 0}{\frac {e^{x+\Delta x}-e^{x}}{\Delta x}}}$ 

${\displaystyle =e^{x}.\lim _{\Delta x\to 0}{\frac {e^{\Delta x-1}}{\Delta x}}}$ 

${\displaystyle =e^{x}.(1)}$ 

${\displaystyle =e^{x}}$ 

五之證：

次第一：${\displaystyle x>0}$ 

${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}(\ln |x|)}$ 

${\displaystyle ={\frac {\mathrm {d} }{\mathrm {d} x}}(\ln(x))}$ 

${\displaystyle =\lim _{\Delta x\to 0}{\frac {\ln(x+\Delta x)-\ln(x)}{\Delta x}}}$ 

${\displaystyle =\lim _{\Delta x\to 0}{\frac {1}{\Delta x}}\ln {\frac {x+\Delta x}{x}}}$ 

${\displaystyle =\lim _{\Delta x\to 0}\ln(1+{\frac {\Delta x}{x}})^{\frac {1}{\Delta x}}}$ 

設 ${\displaystyle n={\frac {x}{\Delta x}}}$  ，則若 ${\displaystyle \Delta x\to 0}$  ， ${\displaystyle n\to \infty }$  。

${\displaystyle \lim _{\Delta x\to 0}\ln(1+{\frac {\Delta x}{x}})^{\frac {1}{\Delta x}}}$ 

${\displaystyle =\ln(\lim _{n\to \infty }(1+{\frac {1}{n}})^{n}))^{\frac {1}{x}}}$ 

${\displaystyle =\ln e^{\frac {1}{x}}}$ 

${\displaystyle ={\frac {1}{x}}}$ 

次第二：${\displaystyle x<0}$ 

${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}(\ln |x|)}$ 

${\displaystyle ={\frac {\mathrm {d} }{\mathrm {d} x}}(\ln(-x))}$ 

${\displaystyle =-{\frac {1}{x}}.{\frac {\mathrm {d} }{\mathrm {d} x}}(-x)}$ 

${\displaystyle =-{\frac {1}{x}}.(-1)}$ 

${\displaystyle ={\frac {1}{x}}}$ 

六之證：

${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}\sin(x)}$ 

${\displaystyle =\lim _{\Delta x\to 0}{\frac {\sin(x+\Delta x)-\sin x}{\Delta x}}}$ 

${\displaystyle =\lim _{\Delta x\to 0}{\frac {2\cos {\frac {2x+\Delta x}{2}}\sin {\frac {\Delta x}{2}}}{\Delta x}}}$ 

${\displaystyle =\lim _{\Delta x\to 0}({\frac {\sin {\frac {\Delta x}{2}}}{\frac {\Delta x}{2}}}.\cos(x+{\frac {\Delta x}{2}}))}$ 

${\displaystyle =1.\cos x}$ 

${\displaystyle =\cos x}$ 

七之證：

${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}\cos(x)}$ 

${\displaystyle =\lim _{\Delta x\to 0}{\frac {\cos(x+\Delta x)-\cos x}{\Delta x}}}$ 

${\displaystyle =\lim _{\Delta x\to 0}{\frac {-2\sin {\frac {2x+\Delta x}{2}}\sin {\frac {\Delta x}{2}}}{\Delta x}}}$ 

${\displaystyle =-\lim _{\Delta x\to 0}({\frac {\sin {\frac {\Delta x}{2}}}{\frac {\Delta x}{2}}}.\sin(x+{\frac {\Delta x}{2}}))}$ 

${\displaystyle =-1.\cos x}$ 

${\displaystyle =-\cos x}$ 

八之證：

${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}(\tan x)}$ 

${\displaystyle ={\frac {\mathrm {d} }{\mathrm {d} x}}({\frac {\sin x}{\cos x}})}$ 

${\displaystyle ={\frac {\cos x{\frac {d}{dx}}(\sin x)-\sin x{\frac {d}{dx}}(\cos x)}{\cos ^{2}x}}}$ 

${\displaystyle ={\frac {\cos ^{2}x+\sin ^{2}x}{\cos ^{2}x}}}$ 

${\displaystyle =\sec ^{2}x}$ 

九之證：

${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}(\csc x)}$ 

${\displaystyle ={\frac {\mathrm {d} }{\mathrm {d} x}}{\frac {1}{\sin x}}}$ 

${\displaystyle ={\frac {\sin x{\frac {\mathrm {d} }{\mathrm {d} x}}(1)-1{\frac {\mathrm {d} }{\mathrm {d} x}}(\sin x)}{\sin ^{2}x}}}$ 

${\displaystyle =-{\frac {\cos x}{\sin ^{2}x}}}$ 

${\displaystyle =-\cot x\csc x}$ 

十之證：

${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}(\sec x)}$ 

${\displaystyle ={\frac {\mathrm {d} }{\mathrm {d} x}}{\frac {1}{\cos x}}}$ 

${\displaystyle ={\frac {\cos x{\mathrm {d} }{\mathrm {d} x}(1)-1{\frac {\mathrm {d} }{\mathrm {d} x}}(\cos x)}{\cos ^{2}x}}}$ 

${\displaystyle ={\frac {\sin x}{\cos ^{2}x}}}$ 

${\displaystyle =\sec x\tan x}$ 

十一之證：

${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}(\cot x)}$ 

${\displaystyle ={\frac {\mathrm {d} }{\mathrm {d} x}}{\frac {1}{\tan x}}}$ 

${\displaystyle ={\frac {\tan x{\frac {\mathrm {d} }{\mathrm {d} x}}(1)-1{\frac {\mathrm {d} }{\mathrm {d} x}}(\tan x)}{\tan ^{2}x}}}$ 

${\displaystyle =-{\frac {\sec ^{2}x}{\tan ^{2}x}}}$ 

${\displaystyle =-\csc ^{2}x}$ 

十二之證：

設 ${\displaystyle x=\sin y}$  且 ${\displaystyle -{\frac {\pi }{2}}\leq y\leq {\frac {\pi }{2}}}$  。則：

${\displaystyle x=\sin y}$ 

${\displaystyle 1=\cos y.{\frac {\mathrm {d} y}{\mathrm {d} x}}}$ 

${\displaystyle {\frac {\mathrm {d} y}{\mathrm {d} x}}={\frac {1}{\cos y}}}$ 

${\displaystyle ={\frac {1}{\sqrt {1-\sin ^{2}y}}}}$ 

${\displaystyle ={\frac {1}{\sqrt {1-x^{2}}}}}$ 

十三之證：

${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}\arccos x}$ 

${\displaystyle ={\frac {\mathrm {d} }{\mathrm {d} x}}({\frac {\pi }{2}}-\arcsin x)}$ 

${\displaystyle =-{\frac {1}{\sqrt {1-x^{2}}}}}$ 

十四之證：

設 ${\displaystyle x=\tan y}$  且 ${\displaystyle -{\frac {\pi }{2}}\leq y\leq {\frac {\pi }{2}}}$  。則：

${\displaystyle x=\tan y}$ 

${\displaystyle 1=\sec ^{2}y.{\frac {\mathrm {d} y}{\mathrm {d} x}}}$ 

${\displaystyle {\frac {\mathrm {d} y}{\mathrm {d} x}}}$ 

${\displaystyle =1\sec ^{2}y}$ 

${\displaystyle ={\frac {1}{1+x^{2}}}}$ 

十五之證：

設 ${\displaystyle u=f(x)}$  且 ${\displaystyle v=g(x)}$  。則：

${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}(u\pm v)}$ 

${\displaystyle =\lim _{\Delta x\to 0}{\frac {(f(x+\Delta x)\pm g(x+\Delta x))-(f(x)\pm g(x))}{\Delta x}}}$ 

${\displaystyle =\lim _{\Delta x\to 0}({\frac {f(x+\Delta x)-f(x)}{\Delta x}}\pm {\frac {g(x+\Delta x)-g(x)}{\Delta x}})}$ 

${\displaystyle =f'(x)\pm g'(x)}$ 

${\displaystyle ={\frac {\mathrm {d} u}{\mathrm {d} x}}\pm {\frac {\mathrm {d} v}{\mathrm {d} x}}}$ 

十六之證：

設 ${\displaystyle u=f(x)}$  且 ${\displaystyle v=g(x)}$  。則：

${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}(uv)}$ 

${\displaystyle =\lim _{\Delta x\to 0}{\frac {f(x+\Delta x)g(x+\Delta x)-f(x)g(x)}{\Delta x}}}$ 

${\displaystyle =\lim _{\Delta x\to 0}{\frac {f(x+\Delta x)g(x+\Delta x)-f(x+\Delta x)g(x)+f(x+\Delta x)g(x)-f(x)g(x)}{\Delta x}}}$ 

${\displaystyle =\lim _{\Delta x\to 0}(f(x+\Delta x)({\frac {g(x+\Delta x)-g(x)}{\Delta x}})+g(x)({\frac {f(x+\Delta x)-f(x)}{\Delta x}}))}$ 

${\displaystyle =f(x)g'(x)+g(x)f'(x)}$ 

${\displaystyle =u{\frac {\mathrm {d} v}{\mathrm {d} x}}+v{\frac {\mathrm {d} u}{\mathrm {d} x}}}$ 

十七之證：

設 ${\displaystyle u=f(x)}$  且 ${\displaystyle v=g(x)}$  。則：

${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}{\frac {u}{v}}}$ 

${\displaystyle =\lim _{\Delta x\to 0}{\frac {{\frac {f(x+\Delta x)}{g(x+\Delta x)}}-{\frac {f(x)}{g(x)}}}{\Delta x}}}$ 

${\displaystyle =\lim _{\Delta x\to 0}{\frac {f(x+\Delta x)g(x)-f(x)g(x+\Delta x)}{\Delta xg(x)g(x+\Delta x)}}}$ 

${\displaystyle =\lim _{\Delta x\to 0}({\frac {1}{g(x)g(x+\Delta x)}}.{\frac {f(x+\Delta x)g(x)-f(x)g(x)+f(x)g(x)-f(x)g(x+\Delta x)}{\Delta x}})}$ 

${\displaystyle =\lim _{\Delta x\to 0}({\frac {1}{g(x)g(x+\Delta x)}}.({\frac {f(x+\Delta x)g(x)-f(x)g(x)}{\Delta x}}-{\frac {f(x)g(x+\Delta x)-f(x)g(x)}{\Delta x}}))}$ 

${\displaystyle ={\frac {1}{g(x)^{2}}}.(g(x)f'(x)-f(x)g'(x))}$ 

${\displaystyle ={\frac {v{\frac {\mathrm {d} u}{\mathrm {d} x}}-u{\frac {\mathrm {d} v}{\mathrm {d} x}}}{v^{2}}}}$ 

十八之證：

設 ${\displaystyle y=f(x)}$  且 ${\displaystyle u=g(x)}$  。則：

${\displaystyle {\frac {\mathrm {d} y}{\mathrm {d} x}}}$ 

${\displaystyle =\lim _{\Delta x\to 0}{\frac {f(g(x+\Delta x))-f(g(x))}{\Delta x}}}$ 

${\displaystyle =\lim _{\Delta x\to 0}{\frac {f(g(x+\Delta x))-f(g(x))}{g(x+\Delta x)-g(x)}}{\frac {g(x+\Delta x)-g(x)}{\Delta x}}}$ 

緣當 ${\displaystyle \Delta x\to 0}$  時， ${\displaystyle g(x+\Delta x)-g(x)=(u+\Delta u)-u=\Delta u\to 0}$  ，故：

${\displaystyle \lim _{\Delta x\to 0}{\frac {f(g(x+\Delta x))-f(g(x))}{g(x+\Delta x)-g(x)}}{\frac {g(x+\Delta x)-g(x)}{\Delta x}}}$ 

${\displaystyle =\lim _{\Delta u\to 0}{\frac {f(u+\Delta u)-f(u)}{\Delta u}}\lim _{\Delta x\to 0}{\frac {g(x+\Delta x)-g(x)}{\Delta x}}}$ 

${\displaystyle =f'(u)g'(x)}$ 

${\displaystyle ={\frac {\mathrm {d} y}{\mathrm {d} u}}{\frac {\mathrm {d} u}{\mathrm {d} x}}}$ 

例

求 ${\displaystyle y=x^{4}+\sin(x^{2})-\ln(x)e^{x}+7}$  之導數。

${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}(x^{4}+\sin(x^{2})-\ln(x)e^{x}+7)}$ 

${\displaystyle =4x^{3}+\cos(x^{2}){\frac {d}{dx}}(x^{2})-({\frac {1}{x}}e^{x}+\ln(x)e^{x})+0}$ 

${\displaystyle =4x^{3}+2x\cos(x^{2})-{\frac {e^{x}}{x}}-\ln(x)e^{x}}$ 

二階導數者，導數之導數也，乃函數於某點時斜率之變率，為以極限趨函數斜率之方程所得也。常書二階導數作${\displaystyle {\frac {\mathrm {d} ^{2}y}{\mathrm {d} x^{2}}}}$ ，${\displaystyle {\frac {\mathrm {d} ^{2}}{\mathrm {d} x^{2}}}f(x)}$ ， ${\displaystyle y''}$ ， ${\displaystyle {\ddot {y}}}$ 等。夫${\displaystyle {\frac {\mathrm {d} ^{2}y}{\mathrm {d} x^{2}}}}$ 者，其義緣${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}({\frac {\mathrm {d} y}{\mathrm {d} x}})}$ 也。

至於甚者（${\displaystyle n}$  階導數，${\displaystyle n\in N}$  當 ${\displaystyle n>2}$  時），其義及書同上。（舉一隅，則反三隅也）

例

求 ${\displaystyle y=4x^{33}+58x^{13}+289x^{4}+5x+1}$  之二階導數。

${\displaystyle {\frac {\mathrm {d} ^{2}y}{\mathrm {d} x^{2}}}}$ 

${\displaystyle ={\frac {\mathrm {d} }{\mathrm {d} x}}(132x^{32}+754x^{12}+1156x^{3}+5)}$ 

${\displaystyle =4224x^{31}+9048x^{11}+3468x^{2}}$ 

詳見偏導數

• https://www.mathsisfun.com/calculus/derivatives-introduction.html
• https://www.mathsisfun.com/calculus/derivatives-rules.html
• https://www.mathsisfun.com/calculus/implicit-differentiation.html
• https://www.mathsisfun.com/calculus/second-derivative.html