導數(英文:Derivative)係微積分裡面好重要嘅基礎概念,可以理解成一個函數喺某一個位嘅瞬間變化。 佢嘅定義係: f ′ ( x ) = lim h → 0 f ( x + h ) − f ( x ) h {\displaystyle f'(x)=\lim _{h\to 0}{f(x+h)-f(x) \over h}} 參考極限。 例如 f ( x ) = x 2 {\displaystyle f(x)=x^{2}} ,求佢嘅導函數 f ′ ( x ) {\displaystyle f'(x)} 。 f ′ ( x ) = lim h → 0 ( x + h ) 2 − x 2 h {\displaystyle f'(x)=\lim _{h\to 0}{(x+h)^{2}-x^{2} \over h}} ,展開變咗 lim h → 0 ( x 2 + 2 h x + h 2 ) − x 2 h {\displaystyle \lim _{h\to 0}{(x^{2}+2hx+h^{2})-x^{2} \over h}} ,消除咗個 x 2 {\displaystyle x^{2}} 淨低 lim h → 0 2 h x + h 2 h {\displaystyle \lim _{h\to 0}{2hx+h^{2} \over h}} ,約埋個 h {\displaystyle h} 變成 lim h → 0 { 2 x + h } {\displaystyle \lim _{h\to 0}\{2x+h\}} ,由於 h {\displaystyle h} 趨向 0 {\displaystyle 0} 所以求到 f ′ ( x ) = 2 x {\displaystyle f'(x)=2x} 。 導數可以搵出條線嘅瞬間變率(instantaneous rate of change)。 d d x c = 0 {\displaystyle {d \over dx}\ c=0} , c {\displaystyle c} 係常數。 d d x a x = a {\displaystyle {d \over dx}\ ax=a} , a {\displaystyle a} 係常數。 d d x a x n = a n x n − 1 {\displaystyle {d \over dx}\ ax^{n}=anx^{n-1}} , a , n {\displaystyle a,n} 係常數。 d d x sin ( x ) = cos ( x ) {\displaystyle {d \over dx}\ \sin(x)=\cos(x)} d d x cos ( x ) = − sin ( x ) {\displaystyle {d \over dx}\ \cos(x)=-\sin(x)} d d x tan ( x ) = sec 2 ( x ) {\displaystyle {d \over dx}\ \tan(x)=\sec ^{2}(x)} d d x sec ( x ) = sec ( x ) tan ( x ) {\displaystyle {d \over dx}\ \sec(x)=\sec(x)\tan(x)} d d x csc ( x ) = − csc ( x ) cot ( x ) {\displaystyle {d \over dx}\ \csc(x)=-\csc(x)\cot(x)} d d x cot ( x ) = − csc 2 ( x ) {\displaystyle {d \over dx}\ \cot(x)=-\csc ^{2}(x)} 。睇下三角函數。 d d x e x = e x {\displaystyle {d \over dx}\ e^{x}=e^{x}} 。睇下自然指數。 d d x a x = a x ln ( a ) {\displaystyle {d \over dx}\ a^{x}=a^{x}\ln(a)} d d x ln ( x ) = 1 x {\displaystyle {d \over dx}\ \ln(x)={1 \over x}} d d x log a ( x ) = 1 x ln ( a ) {\displaystyle {d \over dx}\ \log _{a}(x)={1 \over x\ln(a)}} d d x sin − 1 ( x ) = 1 1 − x 2 {\displaystyle {d \over dx}\ \sin ^{-1}(x)={\frac {1}{\sqrt {1-x^{2}}}}} d d x cos − 1 ( x ) = − 1 1 − x 2 {\displaystyle {d \over dx}\ \cos ^{-1}(x)=-{\frac {1}{\sqrt {1-x^{2}}}}} d d x tan − 1 ( x ) = 1 1 + x 2 {\displaystyle {d \over dx}\ \tan ^{-1}(x)={\frac {1}{1+x^{2}}}} d d x erf ( x ) = 2 π e − x 2 {\displaystyle {d \over dx}\ \operatorname {erf} (x)={\frac {2}{\sqrt {\pi }}}e^{-x^{2}}} 加法: [ f ( x ) + g ( x ) ] ′ = f ′ ( x ) + g ′ ( x ) {\displaystyle [\ f(x)+g(x)\ ]'=f'(x)+g'(x)} 乘法: [ f ( x ) g ( x ) ] ′ = f ′ ( x ) g ( x ) + f ( x ) g ′ ( x ) {\displaystyle [\ f(x)\ g(x)\ ]'=f'(x)\ g(x)+f(x)\ g'(x)} 證明:設 y = f g {\displaystyle y=fg} ln ( y ) = ln ( f g ) {\displaystyle \ln(y)=\ln(fg)} ln ( y ) = ln ( f ) + ln ( g ) {\displaystyle \ln(y)=\ln(f)+\ln(g)} (對數律) [ ln ( y ) ] ′ = [ ln ( f ) + ln ( g ) ] ′ {\displaystyle [\ln(y)]'=[\ln(f)+\ln(g)]'} y ′ y = f ′ f + g ′ g {\displaystyle {y' \over y}={f' \over f}+{g' \over g}} y ′ y = f ′ g + f g ′ f g {\displaystyle {y' \over y}={{f'g+fg'} \over fg}} ( f g ) ′ f g = f ′ g + f g ′ f g {\displaystyle {(fg)' \over fg}={{f'g+fg'} \over fg}} ( f g ) ′ = f ′ g + f g ′ {\displaystyle (fg)'=f'g+fg'} 除法: ( f ( x ) g ( x ) ) ′ = f ′ ( x ) g ( x ) − f ( x ) g ′ ( x ) g ( x ) 2 {\displaystyle ({\frac {f(x)}{g(x)}})'={\frac {f'(x)g(x)-f(x)g'(x)}{g(x)^{2}}}} 連鎖律: d y d x = d y d u d u d x {\displaystyle {dy \over dx}={dy \over du}{du \over dx}} 冪函數嘅導數同ln(x)嘅關係: 對冪函數求導,指數部分會降低1次,如 d d x x 3 = 3 x 2 {\displaystyle {\frac {d}{dx}}x^{3}=3x^{2}} d d x x 2 = 2 x {\displaystyle {\frac {d}{dx}}x^{2}=2x} d d x x = 1 {\displaystyle {\frac {d}{dx}}x=1} (呢度可以當係) x 0 {\displaystyle x^{0}} 但對常數繼續求導唔會變成 -1 次方,因為0乘任何數都係0,即 d d x x 0 = 0 x − 1 = 0 {\displaystyle {\frac {d}{dx}}x^{0}=0x^{-1}=0} 反過嚟,從積分方向考慮 ∫ x − 3 d x = x − 2 − 2 + C 3 {\displaystyle \int x^{-3}dx={\frac {x^{-2}}{-2}}+C_{3}} ∫ x − 2 d x = x − 1 − 1 + C 2 {\displaystyle \int x^{-2}dx={\frac {x^{-1}}{-1}}+C_{2}} ∫ x − 1 d x = x 0 0 + C 1 {\displaystyle \int x^{-1}dx={\frac {x^{0}}{0}}+C_{1}} ,出現0分母,所以唔會變成0之方,正確係 ∫ x − 1 d x = l n ( x ) + C 0 {\displaystyle \int x^{-1}dx=ln(x)+C_{0}} 即冪函數似乎同自然對數函數ln(x)有密切關係。 其實原因好簡單,ln(x)嘅極限式就有一部分係由冪函數組成: l n ( x ) = lim n − > 0 x n − 1 n {\displaystyle ln(x)=\lim _{n->0}{\frac {x^{n}-1}{n}}} 可以當成 x 0 − 1 0 = x 0 0 − 1 0 {\displaystyle {\frac {x^{0}-1}{0}}={\frac {x^{0}}{0}}-{\frac {1}{0}}} 同上面嘅積分式一樣。 可以用極限式求導: d d x l n ( x ) {\displaystyle {\frac {d}{dx}}ln(x)} = d d x lim n → 0 x n − 1 n {\displaystyle ={\frac {d}{dx}}\lim _{n\to 0}{\frac {x^{n}-1}{n}}} = lim n → 0 d d x x n − 1 n {\displaystyle =\lim _{n\to 0}{\frac {d}{dx}}{\frac {x^{n}-1}{n}}} = lim n → 0 n x n − 1 n {\displaystyle =\lim _{n\to 0}{\frac {nx^{n-1}}{n}}} = lim n → 0 x n − 1 {\displaystyle =\lim _{n\to 0}x^{n-1}} = x − 1 {\displaystyle =x^{-1}} 結果一致。 Wikiwand in your browser!Seamless Wikipedia browsing. 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