Знаходження похідної є найважливішою операцією у диференціальному численні .
У цій статті наведені правила диференціювання та список похідних основних функцій , яких достатньо для диференціювання будь-якої елементарної функції.
У нижчеподаних формулах
x
{\displaystyle x}
f
{\displaystyle f}
u
{\displaystyle u}
v
{\displaystyle v}
c
{\displaystyle c}
(
c
⋅ ⋅ -->
f
)
′ ′ -->
=
c
⋅ ⋅ -->
f
′ ′ -->
{\displaystyle \left({c\cdot f}\right)^{\prime }=c\cdot f^{\prime }}
c
=
const
{\displaystyle c={\text{const}}\,}
(
u
+
v
)
′
=
u
′
+
v
′
{\displaystyle \left({u+v}\right)'=u'+v'}
(
u
− − -->
v
)
′
=
u
′
− − -->
v
′
{\displaystyle \left({u-v}\right)'=u'-v'}
(
u
⋅ ⋅ -->
v
)
′
=
u
′
v
+
u
v
′
{\displaystyle \left({u\cdot v}\right)'={u'v+uv'}}
(
u
v
)
′
=
u
′
v
− − -->
u
v
′
v
2
,
v
≠ ≠ -->
0
{\displaystyle \left({u \over v}\right)'={{u'v-uv'} \over {v^{2}}}\,,\qquad v\neq 0}
(
u
(
v
(
x
)
)
)
′
=
u
v
′
(
v
)
⋅ ⋅ -->
v
x
′
(
x
)
{\displaystyle {\Big (}{u{\big (}v(x){\big )}}{\Big )}'={u_{v}'(v)\cdot v_{x}'(x)}}
(
f
− − -->
1
)
′
=
1
f
′
∘ ∘ -->
f
− − -->
1
{\displaystyle (f^{-1})'={\frac {1}{f'\circ f^{-1}}}}
(
c
)
′
=
0
{\displaystyle \left(c\right)'=0}
(
x
)
′
=
1
{\displaystyle \left(x\right)'=1}
(
c
x
)
′
=
c
{\displaystyle \left(cx\right)'=c}
|
x
|
′
=
x
|
x
|
=
sgn
-->
x
,
x
≠ ≠ -->
0
{\displaystyle |x|'={x \over |x|}=\operatorname {sgn} x,\qquad x\neq 0}
(
x
c
)
′
=
c
x
c
− − -->
1
{\displaystyle \left(x^{c}\right)'=cx^{c-1}}
(
x
c
)
{\displaystyle \left(x^{c}\right)}
(
x
c
− − -->
1
)
{\displaystyle \left(x^{c-1}\right)}
Зокрема:
(
1
x
c
)
′
=
d
d
x
(
x
− − -->
c
)
=
− − -->
c
x
c
+
1
{\displaystyle \left({1 \over x^{c}}\right)'={d \over dx}\left(x^{-c}\right)=-{c \over x^{c+1}}}
(
1
x
)
′
=
d
d
x
(
x
− − -->
1
)
=
− − -->
1
x
2
{\displaystyle \left({1 \over x}\right)'={d \over dx}\left(x^{-1}\right)=-{1 \over x^{2}}}
(
x
)
′
=
d
d
x
x
1
2
=
1
2
x
− − -->
1
2
=
1
2
x
,
x
>
0
{\displaystyle \left({\sqrt {x}}\right)'={d \over dx}x^{1 \over 2}={1 \over 2}x^{-{1 \over 2}}={1 \over 2{\sqrt {x}}},\qquad x>0}
(
e
x
)
′
=
e
x
{\displaystyle \left(e^{x}\right)'=e^{x}}
(
ln
-->
x
)
′
=
1
x
{\displaystyle \left(\ln x\right)'={1 \over x}}
(
a
x
)
′
=
a
x
ln
-->
a
,
a
>
0
{\displaystyle \left(a^{x}\right)'={a^{x}\ln a},\quad a>0}
(
log
a
-->
x
)
′
=
1
x
ln
-->
a
,
a
>
0
,
a
≠ ≠ -->
1
{\displaystyle \left(\log _{a}x\right)'={1 \over x\ln a},\quad a>0,a\neq 1}
(
e
a
x
)
′
=
a
e
a
x
{\displaystyle \left(e^{ax}\right)'=ae^{ax}}
Прямих
Обернених
(
sin
-->
x
)
′
=
cos
-->
x
{\displaystyle \left(\sin x\right)'=\cos x}
(
arcsin
-->
x
)
′
=
1
1
− − -->
x
2
{\displaystyle \left(\operatorname {arcsin} x\right)'={1 \over {\sqrt {1-x^{2}}}}}
(
cos
-->
x
)
′
=
− − -->
sin
-->
x
{\displaystyle \left(\cos x\right)'=-\sin x}
(
arccos
-->
x
)
′
=
− − -->
1
1
− − -->
x
2
{\displaystyle \left(\operatorname {arccos} x\right)'={-1 \over {\sqrt {1-x^{2}}}}}
(
tg
-->
x
)
′
=
1
cos
2
-->
x
{\displaystyle \left(\operatorname {tg} x\right)'={1 \over \cos ^{2}x}}
(
arctg
-->
x
)
′
=
1
1
+
x
2
{\displaystyle \left(\operatorname {arctg} x\right)'={1 \over 1+x^{2}}}
(
sec
-->
x
)
′
=
sin
-->
x
cos
2
-->
x
=
tg
-->
x
sec
-->
x
{\displaystyle \left(\sec x\right)'={\sin x \over \cos ^{2}x}=\operatorname {tg} x\sec x}
(
arcsec
-->
x
)
′
=
1
|
x
|
x
2
− − -->
1
{\displaystyle \left(\operatorname {arcsec} x\right)'={1 \over |x|{\sqrt {x^{2}-1}}}}
(
ctg
-->
x
)
′
=
− − -->
1
sin
2
-->
x
{\displaystyle \left(\operatorname {ctg} x\right)'={-1 \over \sin ^{2}x}}
(
arcctg
-->
x
)
′
=
− − -->
1
1
+
x
2
{\displaystyle \left(\operatorname {arcctg} x\right)'={-1 \over 1+x^{2}}}
(
cosec
-->
x
)
′
=
− − -->
cos
-->
x
sin
2
-->
x
=
− − -->
ctg
-->
x
cosec
-->
x
{\displaystyle \left(\operatorname {cosec} x\right)'=-{\cos x \over \sin ^{2}x}=-\operatorname {ctg} x\operatorname {cosec} x}
(
arccosec
-->
x
)
′
=
− − -->
1
|
x
|
x
2
− − -->
1
{\displaystyle \left(\operatorname {arccosec} x\right)'={-1 \over |x|{\sqrt {x^{2}-1}}}}
Прямих
Обернених
(
sh
-->
x
)
′
=
ch
-->
x
{\displaystyle \left(\operatorname {sh} x\right)'=\operatorname {ch} x}
(
arsh
-->
x
)
′
=
1
x
2
+
1
{\displaystyle \left(\operatorname {arsh} x\right)'={1 \over {\sqrt {x^{2}+1}}}}
(
ch
-->
x
)
′
=
sh
-->
x
{\displaystyle \left(\operatorname {ch} x\right)'=\operatorname {sh} x}
(
arch
-->
x
)
′
=
− − -->
1
x
2
− − -->
1
{\displaystyle \left(\operatorname {arch} x\right)'={-1 \over {\sqrt {x^{2}-1}}}}
(
th
-->
x
)
′
=
sch
2
-->
x
{\displaystyle \left(\operatorname {th} x\right)'=\operatorname {sch} ^{2}x}
(
arth
-->
x
)
′
=
1
1
− − -->
x
2
{\displaystyle \left(\operatorname {arth} x\right)'={1 \over 1-x^{2}}}
(
sech
-->
x
)
′
=
− − -->
th
-->
x
sech
-->
x
{\displaystyle \left(\operatorname {sech} x\right)'=-\operatorname {th} x\,\operatorname {sech} x}
(
arsech
-->
x
)
′
=
1
x
1
− − -->
x
2
{\displaystyle \left(\operatorname {arsech} x\right)'={1 \over x{\sqrt {1-x^{2}}}}}
(
cth
-->
x
)
′
=
− − -->
csch
2
-->
x
{\displaystyle \left(\operatorname {cth} x\right)'=-\operatorname {csch} ^{2}x}
(
arcth
-->
x
)
′
=
1
1
− − -->
x
2
{\displaystyle \left(\operatorname {arcth} x\right)'={1 \over 1-x^{2}}}
(
csch
-->
x
)
′
=
− − -->
cth
-->
x
csch
-->
x
{\displaystyle \left(\operatorname {csch} x\right)'=-\operatorname {cth} x\,\operatorname {csch} x}
(
arcsch
-->
x
)
′
=
− − -->
1
|
x
|
1
+
x
2
{\displaystyle \left(\operatorname {arcsch} x\right)'={-1 \over |x|{\sqrt {1+x^{2}}}}}
