Derivêye
Cisse pådje ci n' est co k' on
djermon , dj' ô bén k' el pådje est djusse
sibåtcheye , eyet co trop tene ; et s' divreut ele ecråxhî ene miete. Si vos avoz des cnoxhances so ç' sudjet ci, vos nos ploz aidî, clitchîz sol loyén
« candjî l' pådje » po radjouter des informåcions.
Ene deriveye d' ene fonccion est l' moyén d' trover combén cisse fonccion varîye cwand el cachonêye don elle elaxhe, ès argumint candje.[1]
Tåve des deriveyes
Vocial li lisse des principåles deriveyes : [2]
f
(
x
)
=
n
⟺
f
′
(
x
)
=
0
{\displaystyle f(x)=n\Longleftrightarrow f'(x)=0}
f
(
x
)
=
x
⟺
f
′
(
x
)
=
n
{\displaystyle f(x)=x\Longleftrightarrow f'(x)=n}
f
(
x
)
=
m
x
+
n
⟺
f
′
(
x
)
=
m
{\displaystyle f(x)=mx+n\Longleftrightarrow f'(x)=m}
f
(
x
)
=
x
2
⟺
f
′
(
x
)
=
2
x
{\displaystyle f(x)=x^{2}\Longleftrightarrow f'(x)=2x}
f
(
x
)
=
x
n
⇛
(
n
∈
N
)
⟺
f
′
(
x
)
=
n
x
k
⇛
(
c
w
a
n
d
:
k
=
n
−
1
)
{\displaystyle f(x)=x^{n}\Rrightarrow (n\in \mathrm {N} )\Longleftrightarrow f'(x)=nx^{k}\Rrightarrow (cwand:k=n-1)}
f
(
x
)
=
1
÷
x
⟺
f
′
(
x
)
=
(
−
1
÷
(
x
2
)
)
{\displaystyle f(x)=1\div x\Longleftrightarrow f'(x)=(-1\div (x^{2}))}
f
(
x
)
=
√
x
⟺
f
′
(
x
)
=
(
1
÷
(
2
√
x
)
)
{\displaystyle f(x)=\surd x\Longleftrightarrow f'(x)=(1\div (2\surd x))}
Referinces eyet sourdants
↑ https://www.techno-science.net/definition/6280.html
↑ https://external-content.duckduckgo.com/iu/?u=http%3A%2F%2Fwww.lesbonsprofs.com%2FuploadImages%2Fusers%2F--26.png%3F1535809903644&f=1&nofb=1
The article is a derivative under the Creative Commons Attribution-ShareAlike License .
A link to the original article can be found here and attribution parties here
By using this site, you agree to the Terms of Use . Gpedia ® is a registered trademark of the Cyberajah Pty Ltd