Der er for få eller ingen kildehenvisninger i denne artikel, hvilket er et problem . Du kan hjælpe ved at angive troværdige kilder til de påstande, som fremføres i artiklen. Her er nogle regneregler for differentiation . Det forudsættes at alle funktioner er differentiable og dermed kontinuerte for den givne definitionsmængde :
y
(
x
)
=
k
{\displaystyle y(x)=k}
k
{\displaystyle k}
y
′
(
x
)
=
0
{\displaystyle y'(x)=0}
(1):
y
(
x
)
=
x
⋅
k
{\displaystyle y(x)=x\cdot k}
k
{\displaystyle k}
y
′
(
x
)
=
k
{\displaystyle y'(x)=k}
y
(
x
)
=
k
⋅
x
n
{\displaystyle y(x)=k\cdot x^{n}}
y
′
(
x
)
=
k
⋅
n
⋅
x
n
−
1
{\displaystyle y'(x)=k\cdot n\cdot x^{n-1}}
y
(
x
)
=
1
x
{\displaystyle y(x)={\frac {1}{x}}}
y
′
(
x
)
=
−
1
x
2
{\displaystyle y'(x)=-{\frac {1}{x^{2}}}}
Funktioner der er sammensatte funktioner samt funktioner der er summen, differensen, produktet eller kvotienten af to differentiable funktioner er selv differentiable (med visse, åbenlyse begrænsninger i definitionsmængderne). Differentialkvotienterne kan udregnes efter følgende regler:
(
f
∘
g
)
′
(
x
)
=
(
f
(
g
(
x
)
)
)
′
=
g
′
(
x
)
⋅
f
′
(
g
(
x
)
)
{\displaystyle (f\circ g)'(x)=(f(g(x)))'=g'(x)\cdot f'(g(x))}
(2):
(
f
+
g
)
′
(
x
)
=
f
′
(
x
)
+
g
′
(
x
)
{\displaystyle (f+g)'(x)=f'(x)+g'(x)}
(
f
−
g
)
′
(
x
)
=
f
′
(
x
)
−
g
′
(
x
)
{\displaystyle (f-g)'(x)=f'(x)-g'(x)}
(
f
⋅
g
)
′
(
x
)
=
f
′
(
x
)
⋅
g
(
x
)
+
f
(
x
)
⋅
g
′
(
x
)
{\displaystyle (f\cdot g)'(x)=f'(x)\cdot g(x)+f(x)\cdot g'(x)}
(
1
g
)
′
(
x
)
=
−
g
′
(
x
)
(
g
(
x
)
)
2
{\displaystyle \left({\frac {1}{g}}\right)'(x)=-{\frac {g'(x)}{(g(x))^{2}}}}
(
f
g
)
′
(
x
)
=
f
′
(
x
)
⋅
g
(
x
)
−
f
(
x
)
⋅
g
′
(
x
)
(
g
(
x
)
)
2
=
f
′
(
x
)
⋅
1
g
(
x
)
+
f
(
x
)
⋅
(
−
g
′
(
x
)
(
g
(
x
)
)
2
)
{\displaystyle \left({\frac {f}{g}}\right)'(x)={\frac {f'(x)\cdot g(x)-f(x)\cdot g'(x)}{(g(x))^{2}}}=f'(x)\cdot {\frac {1}{g(x)}}+f(x)\cdot (-{\frac {g'(x)}{(g(x))^{2}}})}
(
f
⋅
h
)
′
(
x
)
=
f
′
(
x
)
⋅
h
(
x
)
+
f
(
x
)
⋅
h
′
(
x
)
{\displaystyle (f\cdot h)'(x)=f'(x)\cdot h(x)+f(x)\cdot h'(x)}
(
1
g
)
′
(
x
)
=
−
g
′
(
x
)
(
g
(
x
)
)
2
{\displaystyle \left({\frac {1}{g}}\right)'(x)=-{\frac {g'(x)}{(g(x))^{2}}}}
Differentiation er linear . (følger af (1) og (2) ) Sinus-funktionen
sin
(
x
)
{\displaystyle \sin(x)}
sin
′
(
x
)
=
cos
x
{\displaystyle \sin '(x)=\cos x}
Cosinus-funktionen
cos
(
x
)
{\displaystyle \cos(x)}
cos
′
(
x
)
=
−
sin
x
{\displaystyle \cos '(x)=-\sin x}
Tangens ,
tan
(
x
)
{\displaystyle \tan(x)}
tan
′
(
x
)
=
1
+
tan
2
(
x
)
{\displaystyle \tan '(x)=1+\tan ^{2}(x)}
Den naturlige eksponentialfunktion ,
e
x
{\displaystyle e^{x}}
e
d
d
x
e
x
=
e
x
{\displaystyle {\frac {d}{dx}}e^{x}=e^{x}}
Eksponentialfunktionen
y
(
x
)
=
a
x
{\displaystyle y(x)=a^{x}}
a
{\displaystyle a}
y
′
(
x
)
=
ln
(
a
)
a
x
{\displaystyle y'(x)=\ln(a)a^{x}}
ln
{\displaystyle \ln }
Den naturlige logaritme ,
ln
(
x
)
{\displaystyle \ln(x)}
ln
′
(
x
)
=
1
x
{\displaystyle \ln '(x)={\frac {1}{x}}}
Differentialkvotienten af summen af to differentiable funktioner
redigér
Sumreglen lader differentialkvotienten af summen af to differentiable funktioner beregnes.
En funktion , f(x), der er givet som en sum:
f
(
x
)
=
v
(
x
)
+
u
(
x
)
{\displaystyle f(x)=v(x)+u(x)}
har differentialkvotienten
f
′
(
x
)
=
v
′
(
x
)
+
u
′
(
x
)
{\displaystyle f'(x)=v'(x)+u'(x)}
Dvs. differentialkvotienten af summen
v
(
x
)
+
u
(
x
)
{\displaystyle v(x)+u(x)}
Først finder vi sekantens hældning , eller differenskvotienten :
Δ
y
Δ
x
=
f
(
x
+
Δ
x
)
−
f
(
x
)
Δ
x
{\displaystyle {\frac {\Delta y}{\Delta x}}={\frac {f(x+\Delta x)-f(x)}{\Delta x}}}
Da f(x)=v(x)+u(x), får vi:
Δ
y
Δ
x
=
v
(
x
+
Δ
x
)
+
u
(
x
+
Δ
x
)
−
(
v
(
x
)
+
u
(
x
)
)
Δ
x
{\displaystyle {\frac {\Delta y}{\Delta x}}={\frac {v(x+\Delta x)+u(x+\Delta x)-(v(x)+u(x))}{\Delta x}}}
Ovenstående ligning kan let omskrives til følgende:
Δ
y
Δ
x
=
v
(
x
+
Δ
x
)
−
v
(
x
)
Δ
x
+
u
(
x
+
Δ
x
)
−
u
(
x
)
Δ
x
{\displaystyle {\frac {\Delta y}{\Delta x}}={\frac {v(x+\Delta x)-v(x)}{\Delta x}}+{\frac {u(x+\Delta x)-u(x)}{\Delta x}}}
Hvis
Δ
x
→
0
{\displaystyle \Delta x\rightarrow 0}
d
d
x
u
(
x
)
+
v
(
x
)
=
v
′
(
x
)
+
u
′
(
x
)
{\displaystyle {\frac {d}{dx}}u(x)+v(x)=v'(x)+u'(x)}
Q
.
E
.
D
.
{\displaystyle Q.E.D.{\frac {}{}}}
Differentialkvotienten af produktet af to differentiable funktioner
redigér
Produktreglen lader differentialkvotienten af produktet af to differentiable funktioner beregnes.
En funktion , f(x), der er givet som et produkt:
f
(
x
)
=
v
(
x
)
⋅
u
(
x
)
{\displaystyle f(x)=v(x)\cdot u(x)}
har differentialkvotienten
f
′
(
x
)
=
v
′
(
x
)
⋅
u
(
x
)
+
v
(
x
)
⋅
u
′
(
x
)
{\displaystyle f'(x)=v'(x)\cdot u(x)+v(x)\cdot u'(x)}
Først findes sekantens hældning , eller differenskvotienten :
Δ
y
Δ
x
=
f
(
x
+
Δ
x
)
−
f
(
x
)
Δ
x
{\displaystyle {\frac {\Delta y}{\Delta x}}={\frac {f(x+\Delta x)-f(x)}{\Delta x}}}
Hvilket, idet f(x) = v(x)u(x), bliver
Δ
y
Δ
x
=
v
(
x
+
Δ
x
)
⋅
u
(
x
+
Δ
x
)
−
v
(
x
)
⋅
u
(
x
)
Δ
x
{\displaystyle {\frac {\Delta y}{\Delta x}}={\frac {v(x+\Delta x)\cdot u(x+\Delta x)-v(x)\cdot u(x)}{\Delta x}}}
Der lægges følgende til differenskvotientens tæller, hvorpå det samme trækkes fra igen. Dette giver 0, således er dette fuldt lovligt.
Δ
y
Δ
x
=
v
(
x
+
Δ
x
)
⋅
u
(
x
+
Δ
x
)
−
v
(
x
)
⋅
u
(
x
)
+
v
(
x
+
Δ
x
)
⋅
u
(
x
)
−
v
(
x
+
Δ
x
)
⋅
u
(
x
)
Δ
x
{\displaystyle {\frac {\Delta y}{\Delta x}}={\frac {v(x+\Delta x)\cdot u(x+\Delta x)-v(x)\cdot u(x)+v(x+\Delta x)\cdot u(x)-v(x+\Delta x)\cdot u(x)}{\Delta x}}}
u(x) og v(x+dx) kan nu sættes uden for parentes, og derefter kan brøkstregen deles op:
Δ
y
Δ
x
=
v
(
x
+
Δ
x
)
[
u
(
x
+
Δ
x
)
−
u
(
x
)
]
+
u
(
x
)
[
v
(
x
+
Δ
x
)
−
v
(
x
)
]
Δ
x
{\displaystyle {\frac {\Delta y}{\Delta x}}={\frac {v(x+\Delta x)[u(x+\Delta x)-u(x)]+u(x)[v(x+\Delta x)-v(x)]}{\Delta x}}}
⇕
{\displaystyle \Updownarrow }
Δ
y
Δ
x
=
v
(
x
+
Δ
x
)
[
u
(
x
+
Δ
x
)
−
u
(
x
)
]
Δ
x
+
u
(
x
)
[
v
(
x
+
Δ
x
)
−
v
(
x
)
]
Δ
x
{\displaystyle {\frac {\Delta y}{\Delta x}}={\frac {v(x+\Delta x)[u(x+\Delta x)-u(x)]}{\Delta x}}+{\frac {u(x)[v(x+\Delta x)-v(x)]}{\Delta x}}}
⇕
{\displaystyle \Updownarrow }
Δ
y
Δ
x
=
v
(
x
+
Δ
x
)
[
u
(
x
+
Δ
x
)
−
u
(
x
)
Δ
x
]
+
u
(
x
)
[
v
(
x
+
Δ
x
)
−
v
(
x
)
Δ
x
]
{\displaystyle {\frac {\Delta y}{\Delta x}}=v(x+\Delta x)\left[{\frac {u(x+\Delta x)-u(x)}{\Delta x}}\right]+u(x)\left[{\frac {v(x+\Delta x)-v(x)}{\Delta x}}\right]}
Differentialkvotienten bliver således:
f
′
(
x
)
=
lim
Δ
x
→
0
(
Δ
y
Δ
x
)
{\displaystyle f'(x)=\lim _{\Delta x\to 0}\left({\frac {\Delta y}{\Delta x}}\right)}
Hvilket i det generelle tilfælde er:
f
′
(
x
)
=
lim
Δ
x
→
0
(
v
(
x
+
Δ
x
)
[
u
(
x
+
Δ
x
)
−
u
(
x
)
Δ
x
]
+
u
(
x
)
[
v
(
x
+
Δ
x
)
−
v
(
x
)
Δ
x
]
)
{\displaystyle f'(x)=\lim _{\Delta x\to 0}\left(v(x+\Delta x)\left[{\frac {u(x+\Delta x)-u(x)}{\Delta x}}\right]+u(x)\left[{\frac {v(x+\Delta x)-v(x)}{\Delta x}}\right]\right)}
⇕
{\displaystyle \Updownarrow }
f
′
(
x
)
=
lim
Δ
x
→
0
(
v
(
x
+
Δ
x
)
)
⋅
lim
Δ
x
→
0
[
u
(
x
+
Δ
x
)
−
u
(
x
)
Δ
x
]
+
lim
Δ
x
→
0
(
u
(
x
)
)
⋅
lim
Δ
x
→
0
[
v
(
x
+
Δ
x
)
−
v
(
x
)
Δ
x
]
{\displaystyle f'(x)=\lim _{\Delta x\to 0}(v(x+\Delta x))\cdot \lim _{\Delta x\to 0}\left[{\frac {u(x+\Delta x)-u(x)}{\Delta x}}\right]+\lim _{\Delta x\to 0}(u(x))\cdot \lim _{\Delta x\to 0}\left[{\frac {v(x+\Delta x)-v(x)}{\Delta x}}\right]}
Der kan nu ses at dette bliver til; hvis de overordnede fire led tages grænseværdien af:
f
′
(
x
)
=
v
(
x
)
⋅
u
′
(
x
)
+
u
(
x
)
⋅
v
′
(
x
)
{\displaystyle f'(x)=v(x)\cdot u'(x)+u(x)\cdot v'(x){\frac {}{}}}
Umiddelbart ville man ikke tro at
lim
Δ
x
→
0
(
v
(
x
+
Δ
x
)
)
=
v
(
x
)
{\displaystyle \lim _{\Delta x\to 0}(v(x+\Delta x))=v(x)}
d
d
x
u
(
x
)
⋅
v
(
x
)
=
v
′
⋅
u
+
v
⋅
u
′
{\displaystyle {\frac {d}{dx}}u(x)\cdot v(x)=v'\cdot u+v\cdot u'}
Q
.
E
.
D
.
{\displaystyle Q.E.D.{\frac {}{}}}
redigér
Kvotientreglen lader differentialkvotienten af brøken af to differentiable funktioner beregnes.
En funktion , f(x), der er givet som en brøk:
f
(
x
)
=
v
(
x
)
u
(
x
)
{\displaystyle f(x)={\frac {v(x)}{u(x)}}}
har differentialkvotienten
f
′
(
x
)
=
v
′
(
x
)
⋅
u
(
x
)
−
v
(
x
)
⋅
u
′
(
x
)
u
(
x
)
2
{\displaystyle f'(x)={\frac {v'(x)\cdot u(x)-v(x)\cdot u'(x)}{u(x)^{2}}}}
Der er for få eller ingen kildehenvisninger i denne artikel, hvilket er et problem. Du kan hjælpe ved at angive troværdige kilder til de påstande, som fremføres i artiklen.
![{\displaystyle {\frac {\Delta y}{\Delta x}}={\frac {v(x+\Delta x)[u(x+\Delta x)-u(x)]+u(x)[v(x+\Delta x)-v(x)]}{\Delta x}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1addff12d758eafa2ab7b42df4d8d3456d08ee53)
![{\displaystyle {\frac {\Delta y}{\Delta x}}={\frac {v(x+\Delta x)[u(x+\Delta x)-u(x)]}{\Delta x}}+{\frac {u(x)[v(x+\Delta x)-v(x)]}{\Delta x}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ed83a820b9e88ec0c6943394199ea8f8d17b64f3)
![{\displaystyle {\frac {\Delta y}{\Delta x}}=v(x+\Delta x)\left[{\frac {u(x+\Delta x)-u(x)}{\Delta x}}\right]+u(x)\left[{\frac {v(x+\Delta x)-v(x)}{\Delta x}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2381708071e17563a41898d179342940fbe5aa00)
![{\displaystyle f'(x)=\lim _{\Delta x\to 0}\left(v(x+\Delta x)\left[{\frac {u(x+\Delta x)-u(x)}{\Delta x}}\right]+u(x)\left[{\frac {v(x+\Delta x)-v(x)}{\Delta x}}\right]\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d4021f833a13d0ec65a702c71333eab98bff4b76)
![{\displaystyle f'(x)=\lim _{\Delta x\to 0}(v(x+\Delta x))\cdot \lim _{\Delta x\to 0}\left[{\frac {u(x+\Delta x)-u(x)}{\Delta x}}\right]+\lim _{\Delta x\to 0}(u(x))\cdot \lim _{\Delta x\to 0}\left[{\frac {v(x+\Delta x)-v(x)}{\Delta x}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7ba36fd3ce24a12cc01f7c3a9c64e39c96885701)
