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Таблица на изводи од елементарни функции и правила за диференцирање

Module by: Liljana Stefanovska. E-mail the author

Summary: Се дава таблицата на изводи од елементарни функции и основните правила за пресметување на извод.

Таблица на изводи од елементарни функции и правила за диференцирање

Преку дефиницијата на извод, за секој тип елементарна функција може да се пресмета изводот. Но пресметувањето на изводот по дефиниција значи негово барање преку гранична вредност, што не секогаш е брз и лесен начин на пресметување. Затоа за секоја елементарна функција е пресметан изводот по дефиниција и се дава таблица на изводи од елементарните функции.

Таблица на изводи од елементарни функции

Table 1: Таблични изводи
y = c , c = const y = c , c = const size 12{y=c,c= ital "const"} {} y ' = 0 y ' = 0 size 12{ { {y}} sup { ' }=0} {}
y = x n y = x n size 12{y=x rSup { size 8{n} } } {} y ' = nx n − 1 y ' = nx n − 1 size 12{ { {y}} sup { ' }= ital "nx" rSup { size 8{n - 1} } } {}
y = e x y = e x size 12{y=e rSup { size 8{x} } } {} y ' = e x y ' = e x size 12{ { {y}} sup { ' }=e rSup { size 8{x} } } {}
y = a x y = a x size 12{y=a rSup { size 8{x} } } {} y ' = a x ln a y ' = a x ln a size 12{ { {y}} sup { ' }=a rSup { size 8{x} } "ln"a} {}
y = ln x y = ln x size 12{y="ln"x} {} y ' = 1 x y ' = 1 x size 12{ { {y}} sup { ' }= { {1} over {x} } } {}
y = log a x y = log a x size 12{y="log" rSub { size 8{a} } x} {} y ' = 1 x ln a y ' = 1 x ln a size 12{ { {y}} sup { ' }= { {1} over {x"ln"a} } } {}
y = sin x y = sin x size 12{y="sin"x} {} y ' = cos x y ' = cos x size 12{ { {y}} sup { ' }="cos"x} {} y ' = cos x y ' = cos x size 12{ { {y}} sup { ' }="cos"x} {}
y = cos x y = cos x size 12{y="cos"x} {} y ' = − sin x y ' = − sin x size 12{ { {y}} sup { ' }= - "sin"x} {}
y = tg x y = tg x size 12{y="tg"`x} {} y ' = 1 cos 2 x y ' = 1 cos 2 x size 12{ { {y}} sup { ' }= { {1} over {"cos" rSup { size 8{2} } x} } } {}
y = ctg x y = ctg x size 12{y="ctg"`x} {} y ' = − 1 sin 2 x y ' = − 1 sin 2 x size 12{ { {y}} sup { ' }= - { {1} over {"sin" rSup { size 8{2} } x} } } {}
y = arcsin x y = arcsin x size 12{y="arcsin"x} {} y ' = 1 1 − x 2 y ' = 1 1 − x 2 size 12{ { {y}} sup { ' }= { {1} over { sqrt {1 - x rSup { size 8{2} } } } } } {}
y = arccos x y = arccos x size 12{y="arccos"x} {} y ' = − 1 1 − x 2 y ' = − 1 1 − x 2 size 12{ { {y}} sup { ' }= - { {1} over { sqrt {1 - x rSup { size 8{2} } } } } } {}
y = arctg x y = arctg x size 12{y="arctg"`x} {} y ' = 1 1 + x 2 y ' = 1 1 + x 2 size 12{ { {y}} sup { ' }= { {1} over {1+x rSup { size 8{2} } } } } {}
y = arcctg x y = arcctg x size 12{y="arcctg"`x} {} y ' = − 1 1 + x 2 y ' = − 1 1 + x 2 size 12{ { {y}} sup { ' }= - { {1} over {1+x rSup { size 8{2} } } } } {}

Правила за диференцирање

Ќе ги докажеме правилата за пресметување на збир (разлика), производ и количник од диференцијабилни функции.

Нека и f(x)f(x) size 12{f \( x \) } {} и g(x)g(x) size 12{g \( x \) } {} се две диференцијабилни функции што значи дека постојат изводите

lim Δx → 0 f ( x + Δx ) − f ( x ) Δx = f ' ( x ) lim Δx → 0 f ( x + Δx ) − f ( x ) Δx = f ' ( x ) size 12{ {"lim"} cSub { size 8{Δx rightarrow 0} } { {f \( x+Δx \) - f \( x \) } over {Δx} } = { {f}} sup { ' } \( x \) } {}

и

limΔx→0g(x+Δx)−g(x)Δx=g'(x)limΔx→0g(x+Δx)−g(x)Δx=g'(x) size 12{ {"lim"} cSub { size 8{Δx rightarrow 0} } { {g \( x+Δx \) - g \( x \) } over {Δx} } = { {g}} sup { ' } \( x \) } {}.

Извод од збир на функции

Теорема 1.

Да се докаже дека

f ( x ) + g ( x ) ′ = f ' ( x ) + g ' ( x ) . f ( x ) + g ( x ) ′ = f ' ( x ) + g ' ( x ) . size 12{ left [f \( x \) +g \( x \) right ] rSup { size 8{′} } = { {f}} sup { ' } \( x \) + { {g}} sup { ' } \( x \) "." } {}
(1)
Доказ.

Теоремата ќе се докаже со пресметување на изводот по дефиниција.

Ако

y=f(x)+g(x)y=f(x)+g(x) size 12{y=f \( x \) +g \( x \) } {},

тогаш

Δy=f(x+Δx)−f(x)+g(x+Δx)−g(x)Δy=f(x+Δx)−f(x)+g(x+Δx)−g(x) size 12{Δy=f \( x+Δx \) - f \( x \) +g \( x+Δx \) - g \( x \) } {}.

Пресметувајќи го изводот по дефиниција се добива

y ' = f ( x ) + g ( x ) ′ = y ' = f ( x ) + g ( x ) ′ = size 12{ { {y}} sup { ' }= left [f \( x \) +g \( x \) right ] rSup { size 8{′} } ={}} {}
(2)
= lim Δx → 0 Δy Δx = lim Δx → 0 f ( x + Δx ) − f ( x ) + g ( x + Δx ) − g ( x ) Δx = = lim Δx → 0 Δy Δx = lim Δx → 0 f ( x + Δx ) − f ( x ) + g ( x + Δx ) − g ( x ) Δx = size 12{ {}= {"lim"} cSub { size 8{Δx rightarrow 0} } { {Δy} over {Δx} } = {"lim"} cSub { size 8{Δx rightarrow 0} } { {f \( x+Δx \) - f \( x \) +g \( x+Δx \) - g \( x \) } over {Δx} } ={}} {}
(3)
lim Δx → 0 f ( x + Δx ) − f ( x ) Δx + lim Δx → 0 g ( x + Δx ) − g ( x ) Δx = f ' ( x ) + g ' ( x ) lim Δx → 0 f ( x + Δx ) − f ( x ) Δx + lim Δx → 0 g ( x + Δx ) − g ( x ) Δx = f ' ( x ) + g ' ( x ) size 12{ {"lim"} cSub { size 8{Δx rightarrow 0} } { {f \( x+Δx \) - f \( x \) } over {Δx} } + {"lim"} cSub { size 8{Δx rightarrow 0} } { {g \( x+Δx \) - g \( x \) } over {Δx} } = { {f}} sup { ' } \( x \) + { {g}} sup { ' } \( x \) } {}
(4)

што значи дека

f(x)+g(x)′=f'(x)+g'(x)f(x)+g(x)′=f'(x)+g'(x) size 12{ left [f \( x \) +g \( x \) right ] rSup { size 8{′} } = { {f}} sup { ' } \( x \) + { {g}} sup { ' } \( x \) } {}.

Example 1: ПРИМЕР 1.

Да се пресмета изводот на функцијата y=sinx+x4+x.y=sinx+x4+x. size 12{y="sin"x+x rSup { size 8{4} } + sqrt {x} "." } {}

РЕШЕНИЕ:

y ' = ( sin x ) ' + ( x 4 ) ' + ( x 1 / 2 ) ' = cos x + 4x 3 + 1 2 x − 1 / 2 = cos x + 4x 3 + 1 2 x . y ' = ( sin x ) ' + ( x 4 ) ' + ( x 1 / 2 ) ' = cos x + 4x 3 + 1 2 x − 1 / 2 = cos x + 4x 3 + 1 2 x . alignl { stack { size 12{ { {y}} sup { ' }= \( "sin"x { { \) }} sup { ' }+ \( x rSup { size 8{4} } { { \) }} sup { ' }+ \( x rSup { size 8{1/2} } { { \) }} sup { ' }={}} {} # ="cos"x+4x rSup { size 8{3} } + { {1} over {2} } x rSup { size 8{ - 1/2} } ={} {} # ="cos"x+4x rSup { size 8{3} } + { {1} over {2 sqrt {x} } } "." {} } } {}

На сосема ист начин се докажува дека и f(x)−g(x)′=f'(x)−g'(x)f(x)−g(x)′=f'(x)−g'(x) size 12{ left [f \( x \) - g \( x \) right ] rSup { size 8{′} } = { {f}} sup { ' } \( x \) - { {g}} sup { ' } \( x \) } {}.

Извод од производ на функции

Теорема 2.

Да се докаже дека

f(x)g(x)′=f'(x)g(x)+f(x)g'(x)f(x)g(x)′=f'(x)g(x)+f(x)g'(x) size 12{ left [f \( x \) g \( x \) right ] rSup { size 8{′} } = { {f}} sup { ' } \( x \) g \( x \) +f \( x \) { {g}} sup { ' } \( x \) } {}.

Доказ.

Нека

y=f(x)g(x)y=f(x)g(x) size 12{y=f \( x \) g \( x \) } {},

тогаш нараснувањето на функцијата е

Δy = f ( x + Δx ) g ( x + Δx ) − f ( x ) g ( x ) Δy = f ( x + Δx ) g ( x + Δx ) − f ( x ) g ( x ) size 12{Δy=f \( x+Δx \) g \( x+Δx \) - f \( x \) g \( x \) } {}
(5)

и додавајќи го и одземајќи го од десната страна изразот

f ( x ) g ( x + Δx ) f ( x ) g ( x + Δx ) size 12{f \( x \) g \( x+Δx \) } {}
(6)

се добива

Δy = f ( x + Δx ) g ( x + Δx ) − f ( x ) g ( x ) + Δy = f ( x + Δx ) g ( x + Δx ) − f ( x ) g ( x ) + size 12{Δy=f \( x+Δx \) g \( x+Δx \) - f \( x \) g \( x \) +{}} {}

+ f ( x ) g ( x + Δx ) − f ( x ) g ( x + Δx ) = + f ( x ) g ( x + Δx ) − f ( x ) g ( x + Δx ) = size 12{+f \( x \) g \( x+Δx \) - f \( x \) g \( x+Δx \) ={}} {}

=f(x+Δx)−f(x)g(x+Δx)+f(x)g(x+Δx)−g(x)=f(x+Δx)−f(x)g(x+Δx)+f(x)g(x+Δx)−g(x) size 12{ {}= left [f \( x+Δx \) - f \( x \) right ]g \( x+Δx \) +f \( x \) left [g \( x+Δx \) - g \( x \) right ]} {}.

Со примена на дефиницијата за извод на функцијата y=f(x)g(x)y=f(x)g(x) size 12{y=f \( x \) g \( x \) } {} се добива

y ' = f ( x ) g ( x ) ′ = y ' = f ( x ) g ( x ) ′ = size 12{ { {y}} sup { ' }= left [f \( x \) g \( x \) right ] rSup { size 8{′} } ={}} {}

= lim Δx → 0 Δy Δx = = lim Δx → 0 Δy Δx = size 12{ {}= { {"lim"} cSub { size 8{Δx rightarrow 0} } { {Δy} over {Δx} } } cSub {} ={}} {}

= lim Δx → 0 f ( x + Δx ) − f ( x ) Δx lim Δx → 0 g ( x + Δx ) + f ( x ) lim Δx → 0 g ( x + Δx ) − g ( x ) Δx = = lim Δx → 0 f ( x + Δx ) − f ( x ) Δx lim Δx → 0 g ( x + Δx ) + f ( x ) lim Δx → 0 g ( x + Δx ) − g ( x ) Δx = size 12{ {}= {"lim"} cSub { size 8{Δx rightarrow 0} } { {f \( x+Δx \) - f \( x \) } over {Δx} } {"lim"} cSub { size 8{Δx rightarrow 0} } g \( x+Δx \) +f \( x \) {"lim"} cSub { size 8{Δx rightarrow 0} } { {g \( x+Δx \) - g \( x \) } over {Δx} } ={}} {}

= f ' ( x ) g ( x ) + f ( x ) g ' ( x ) = f ' ( x ) g ( x ) + f ( x ) g ' ( x ) size 12{ {}= { {f}} sup { ' } \( x \) g \( x \) +f \( x \) { {g}} sup { ' } \( x \) } {}

или

f(x)g(x)′=f'(x)g(x)+f(x)g'(x)f(x)g(x)′=f'(x)g(x)+f(x)g'(x) size 12{ left [f \( x \) g \( x \) right ] rSup { size 8{′} } = { {f}} sup { ' } \( x \) g \( x \) +f \( x \) { {g}} sup { ' } \( x \) } {}.

Последица.

cf ( x ) ′ = c f ' ( x ) , c − const . cf ( x ) ′ = c f ' ( x ) , c − const . size 12{ left [ ital "cf" \( x \) right ] rSup { size 8{′} } =c { {f}} sup { ' } \( x \) ,c - ital "const" "." } {}
(7)

Example 2: ПРИМЕР 2.

Да се пресмета изводот на функцијата y=5lnx−x2ex.y=5lnx−x2ex. size 12{y=5"ln"x - x rSup { size 8{2} } e rSup { size 8{x} } "." } {}

РЕШЕНИЕ:

y ' = 5 ( ln x ) ' − ( x 2 ) ' e x − x 2 ( e x ) ' = 5 x − 2 xe x − x 2 e x . y ' = 5 ( ln x ) ' − ( x 2 ) ' e x − x 2 ( e x ) ' = 5 x − 2 xe x − x 2 e x . size 12{ { {y}} sup { ' }=5 \( "ln"x { { \) }} sup { ' } - \( x rSup { size 8{2} } { { \) }} sup { ' }e rSup { size 8{x} } - x rSup { size 8{2} } \( e rSup { size 8{x} } { { \) }} sup { ' }= { {5} over {x} } - 2 ital "xe" rSup { size 8{x} } - x rSup { size 8{2} } e rSup { size 8{x} } "." } {}

Извод од количник на функции

Теорема 3.

Да се докаже дека

f ( x ) g ( x ) ′ = f ' ( x ) g ( x ) − f ( x ) g ' ( x ) g 2 ( x ) . f ( x ) g ( x ) ′ = f ' ( x ) g ( x ) − f ( x ) g ' ( x ) g 2 ( x ) . size 12{ left [ { {f \( x \) } over {g \( x \) } } right ] rSup { size 8{′} } = { { { {f}} sup { ' } \( x \) g \( x \) - f \( x \) { {g}} sup { ' } \( x \) } over {g rSup { size 8{2} } \( x \) } } "." } {}
(8)
Доказ.

Нека

y=f(x)g(x)y=f(x)g(x) size 12{y= { {f \( x \) } over {g \( x \) } } } {},

тогаш

Δy=f(x+Δx)g(x+Δx)−f(x)g(x)=f(x+Δx)g(x)−f(x)g(x+Δx)g(x+Δx)g(x)Δy=f(x+Δx)g(x+Δx)−f(x)g(x)=f(x+Δx)g(x)−f(x)g(x+Δx)g(x+Δx)g(x) size 12{Δy= { {f \( x+Δx \) } over {g \( x+Δx \) } } - { {f \( x \) } over {g \( x \) } } = { {f \( x+Δx \) g \( x \) - f \( x \) g \( x+Δx \) } over {g \( x+Δx \) g \( x \) } } } {}.

Додавајќи го и одземајќи го од броителот изразот f(x)g(x)f(x)g(x) size 12{f \( x \) g \( x \) } {} се добива

Δy = f ( x + Δx ) g ( x ) − f ( x ) g ( x + Δx ) + f ( x ) g ( x ) − f ( x ) g ( x ) g ( x + Δx ) g ( x ) = Δy = f ( x + Δx ) g ( x ) − f ( x ) g ( x + Δx ) + f ( x ) g ( x ) − f ( x ) g ( x ) g ( x + Δx ) g ( x ) = size 12{Δy= { {f \( x+Δx \) g \( x \) - f \( x \) g \( x+Δx \) +f \( x \) g \( x \) - f \( x \) g \( x \) } over {g \( x+Δx \) g \( x \) } } ={}} {}
(9)
= f ( x + Δx ) − f ( x ) g ( x ) − f ( x ) g ( x + Δx ) − g ( x ) g ( x + Δx ) g ( x ) . = f ( x + Δx ) − f ( x ) g ( x ) − f ( x ) g ( x + Δx ) − g ( x ) g ( x + Δx ) g ( x ) . size 12{ {}= { { left [f \( x+Δx \) - f \( x \) right ]g \( x \) - f \( x \) left [g \( x+Δx \) - g \( x \) right ]} over {g \( x+Δx \) g \( x \) } } "." } {}
(10)

Пресметувајќи го изводот по дефиниција за функцијата y=f(x)g(x)y=f(x)g(x) size 12{y= { {f \( x \) } over {g \( x \) } } } {} се добива

y ' = f ( x ) g ( x ) ′ = y ' = f ( x ) g ( x ) ′ = size 12{ { {y}} sup { ' }= left [ { {f \( x \) } over {g \( x \) } } right ] rSup { size 8{′} } ={}} {}
(11)

= lim Δx → 0 Δy Δx = = lim Δx → 0 Δy Δx = size 12{ {}= { {"lim"} cSub { size 8{Δx rightarrow 0} } { {Δy} over {Δx} } } cSub {} ={}} {}

= lim Δx → 0 f ( x + Δx ) − f ( x ) Δx g ( x ) − f ( x ) lim Δx → 0 g ( x + Δx ) − g ( x ) Δx lim Δx → 0 g ( x + Δx ) g ( x ) = = lim Δx → 0 f ( x + Δx ) − f ( x ) Δx g ( x ) − f ( x ) lim Δx → 0 g ( x + Δx ) − g ( x ) Δx lim Δx → 0 g ( x + Δx ) g ( x ) = size 12{ {}= { { {"lim"} cSub { size 8{Δx rightarrow 0} } { {f \( x+Δx \) - f \( x \) } over {Δx} } g \( x \) - f \( x \) {"lim"} cSub { size 8{Δx rightarrow 0} } { {g \( x+Δx \) - g \( x \) } over {Δx} } } over { {"lim"} cSub { size 8{Δx rightarrow 0} } g \( x+Δx \) g \( x \) } } ={}} {}

= f ' ( x ) g ( x ) − f ( x ) g ' ( x ) g 2 ( x ) = f ' ( x ) g ( x ) − f ( x ) g ' ( x ) g 2 ( x ) size 12{ {}= { { { {f}} sup { ' } \( x \) g \( x \) - f \( x \) { {g}} sup { ' } \( x \) } over {g rSup { size 8{2} } \( x \) } } } {}

што докажува дека

f(x)g(x)′=f'(x)g(x)−f(x)g'(x)g2(x)f(x)g(x)′=f'(x)g(x)−f(x)g'(x)g2(x) size 12{ left [ { {f \( x \) } over {g \( x \) } } right ] rSup { size 8{′} } = { { { {f}} sup { ' } \( x \) g \( x \) - f \( x \) { {g}} sup { ' } \( x \) } over {g rSup { size 8{2} } \( x \) } } } {}.

Example 3: ПРИМЕР 3.

Да се пресмета изводот на функцијата y=2x3a2−x2.y=2x3a2−x2. size 12{y= { {2x rSup { size 8{3} } } over {a rSup { size 8{2} } - x rSup { size 8{2} } } } "." } {}

РЕШЕНИЕ:

y ' = ( 2x 3 ) ' ( a 2 − x 2 ) − 2x 3 ( a 2 − x 2 ) ' ( a 2 − x 2 ) 2 = 6x 2 ( a 2 − x 2 ) − 2x 3 ( − 2x ) ( a 2 − x 2 ) 2 = 6x 2 a 2 − 2x 4 ( a 2 − x 2 ) 2 . y ' = ( 2x 3 ) ' ( a 2 − x 2 ) − 2x 3 ( a 2 − x 2 ) ' ( a 2 − x 2 ) 2 = 6x 2 ( a 2 − x 2 ) − 2x 3 ( − 2x ) ( a 2 − x 2 ) 2 = 6x 2 a 2 − 2x 4 ( a 2 − x 2 ) 2 . size 12{ { {y}} sup { ' }= { { \( 2x rSup { size 8{3} } { { \) }} sup { ' } \( a rSup { size 8{2} } - x rSup { size 8{2} } \) - 2x rSup { size 8{3} } \( a rSup { size 8{2} } - x rSup { size 8{2} } { { \) }} sup { ' }} over { \( a rSup { size 8{2} } - x rSup { size 8{2} } \) rSup { size 8{2} } } } = { {6x rSup { size 8{2} } \( a rSup { size 8{2} } - x rSup { size 8{2} } \) - 2x rSup { size 8{3} } \( - 2x \) } over { \( a rSup { size 8{2} } - x rSup { size 8{2} } \) rSup { size 8{2} } } } = { {6x rSup { size 8{2} } a rSup { size 8{2} } - 2x rSup { size 8{4} } } over { \( a rSup { size 8{2} } - x rSup { size 8{2} } \) rSup { size 8{2} } } } "." } {}

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Lens makers point to materials (modules and collections), creating a guide that includes their own comments and descriptive tags about the content.

Who can create a lens?

Any individual member, a community, or a respected organization.

What are tags? tag icon

Tags are descriptors added by lens makers to help label content, attaching a vocabulary that is meaningful in the context of the lens.

| External bookmarks

Reuse / Edit:

Reuse or edit collection (?)

Check out and edit

If you have permission to edit this content, using the "Reuse / Edit" action will allow you to check the content out into your Personal Workspace or a shared Workgroup and then make your edits.

Derive a copy

If you don't have permission to edit the content, you can still use "Reuse / Edit" to adapt the content by creating a derived copy of it and then editing and publishing the copy.

| Reuse or edit module (?)

Check out and edit

If you have permission to edit this content, using the "Reuse / Edit" action will allow you to check the content out into your Personal Workspace or a shared Workgroup and then make your edits.

Derive a copy

If you don't have permission to edit the content, you can still use "Reuse / Edit" to adapt the content by creating a derived copy of it and then editing and publishing the copy.