Вычисление производной — операция в дифференциальном исчислении . Эта статья содержит список формул для нахождения производных от некоторых функций.
В этих формулах
f
{\displaystyle f}
g
{\displaystyle g}
вещественной переменной , а
c
{\displaystyle c}
элементарной функции .
d
d
x
c
=
0
{\displaystyle {d \over dx}c=0}
d
d
x
x
=
1
{\displaystyle {d \over dx}x=1}
d
d
x
c
x
=
c
{\displaystyle {d \over dx}cx=c}
(
c
x
)
′
=
c
x
′
=
c
{\displaystyle (cx)'=cx'=c}
d
d
x
x
c
=
c
x
c
−
1
,
{\displaystyle {d \over dx}x^{c}=cx^{c-1},}
x
c
{\displaystyle x^{c}}
c
x
c
−
1
{\displaystyle cx^{c-1}}
c
≠
0
{\displaystyle c\neq 0}
(
x
+
h
)
c
=
x
c
+
(
x
c
)
′
h
+
o
(
h
)
{\displaystyle (x+h)^{c}=x^{c}+(x^{c})'h+o(h)}
(
x
+
h
)
c
−
x
c
=
(
x
c
)
′
h
+
o
(
h
)
{\displaystyle (x+h)^{c}-x^{c}=(x^{c})'h+o(h)}
∑
k
=
0
c
(
c
k
)
x
c
−
k
h
k
−
x
c
=
(
x
c
)
′
h
+
o
(
h
)
{\displaystyle \sum _{k=0}^{c}{c \choose k}x^{c-k}h^{k}-x^{c}=(x^{c})'h+o(h)}
x
c
+
c
x
c
−
1
h
+
∑
k
=
2
c
(
c
k
)
x
c
−
k
h
k
−
x
c
=
(
x
c
)
′
h
+
o
(
h
)
{\displaystyle x^{c}+cx^{c-1}h+\sum _{k=2}^{c}{c \choose k}x^{c-k}h^{k}-x^{c}=(x^{c})'h+o(h)}
c
x
c
−
1
h
+
∑
k
=
2
c
(
c
k
)
x
c
−
k
h
k
=
(
x
c
)
′
h
+
o
(
h
)
{\displaystyle cx^{c-1}h+\sum _{k=2}^{c}{c \choose k}x^{c-k}h^{k}=(x^{c})'h+o(h)}
c
x
c
−
1
h
+
o
(
h
)
=
(
x
c
)
′
h
+
o
(
h
)
{\displaystyle cx^{c-1}h+o(h)=(x^{c})'h+o(h)}
lim
h
→
0
(
c
x
c
−
1
+
o
(
h
)
h
)
=
lim
h
→
0
(
(
x
c
)
′
+
o
(
h
)
h
)
{\displaystyle \lim _{h\rightarrow 0}(cx^{c-1}+{\frac {o(h)}{h}})=\lim _{h\rightarrow 0}((x^{c})'+{\frac {o(h)}{h}})}
c
x
c
−
1
=
(
x
c
)
′
{\displaystyle cx^{c-1}=(x^{c})'}
d
d
x
|
x
|
=
x
|
x
|
=
sgn
x
,
x
≠
0
{\displaystyle {d \over dx}|x|={x \over |x|}=\operatorname {sgn} x,\qquad x\neq 0}
Так как
|
x
|
=
x
2
{\displaystyle |x|={\sqrt {x^{2}}}}
g
(
x
)
=
x
2
,
h
(
x
)
=
x
{\displaystyle g(x)=x^{2},\quad h(x)={\sqrt {x}}}
f
(
x
)
=
h
(
g
(
x
)
)
=
x
2
=
|
x
|
{\displaystyle f(x)=h(g(x))={\sqrt {x^{2}}}=|x|}
Тогда
f
′
(
x
)
=
h
′
(
g
(
x
)
)
⋅
g
′
(
x
)
=
1
2
x
2
⋅
2
x
=
x
x
2
=
x
|
x
|
{\displaystyle f'(x)=h'(g(x))\cdot g'(x)={\frac {1}{2{\sqrt {x^{2}}}}}\cdot 2x={\frac {x}{\sqrt {x^{2}}}}={\frac {x}{|x|}}}
d
d
x
(
1
x
)
=
d
d
x
(
x
−
1
)
=
−
x
−
2
=
−
1
x
2
{\displaystyle {d \over dx}\left({1 \over x}\right)={d \over dx}\left(x^{-1}\right)=-x^{-2}=-{1 \over x^{2}}}
d
d
x
(
1
x
c
)
=
d
d
x
(
x
−
c
)
=
−
c
x
c
+
1
{\displaystyle {d \over dx}\left({1 \over x^{c}}\right)={d \over dx}\left(x^{-c}\right)=-{c \over x^{c+1}}}
d
d
x
x
=
d
d
x
x
1
2
=
1
2
x
−
1
2
=
1
2
x
,
x
>
0
{\displaystyle {d \over dx}{\sqrt {x}}={d \over dx}x^{1 \over 2}={1 \over 2}x^{-{1 \over 2}}={1 \over 2{\sqrt {x}}},\qquad x>0}
d
d
x
x
n
=
d
d
x
x
1
n
=
1
n
x
1
−
n
n
=
1
n
⋅
x
n
−
1
n
{\displaystyle {d \over dx}{\sqrt[{n}]{x}}={d \over dx}x^{1 \over n}={1 \over n}x^{1-n \over n}={\frac {1}{n\cdot {\sqrt[{n}]{x^{n-1}}}}}}
d
d
x
c
x
=
c
x
ln
c
,
c
>
0
{\displaystyle {d \over dx}c^{x}={c^{x}\ln c},\qquad c>0}
d
d
x
c
x
=
d
d
x
e
x
ln
c
=
e
x
ln
c
ln
c
=
c
x
ln
c
{\displaystyle {d \over dx}c^{x}={d \over dx}e^{x\ln c}=e^{x\ln c}\ln c=c^{x}\ln c}
d
d
x
e
x
=
e
x
{\displaystyle {d \over dx}e^{x}=e^{x}}
d
d
x
e
f
(
x
)
=
f
′
(
x
)
e
f
(
x
)
{\displaystyle {d \over dx}e^{f(x)}=f'(x)e^{f(x)}}
d
d
x
ln
x
=
1
x
{\displaystyle {d \over dx}\ln x={1 \over x}}
d
d
x
log
a
x
=
log
a
e
x
=
1
x
ln
a
{\displaystyle {d \over dx}\log _{a}x={\frac {\log _{a}e}{x}}={\frac {1}{x\ln a}}}
l
o
g
a
(
x
+
h
)
=
l
o
g
a
x
+
(
l
o
g
a
x
)
′
h
+
o
(
h
)
{\displaystyle log_{a}(x+h)=log_{a}x+(log_{a}x)'h+o(h)}
l
o
g
a
(
x
+
h
)
−
l
o
g
a
x
=
(
l
o
g
a
x
)
′
h
+
o
(
h
)
{\displaystyle log_{a}(x+h)-log_{a}x=(log_{a}x)'h+o(h)}
l
o
g
a
(
1
+
h
x
)
=
(
l
o
g
a
x
)
′
h
+
o
(
h
)
{\displaystyle log_{a}(1+{\frac {h}{x}})=(log_{a}x)'h+o(h)}
l
o
g
a
e
h
x
=
(
l
o
g
a
x
)
′
h
+
o
(
h
)
{\displaystyle log_{a}e{\frac {h}{x}}=(log_{a}x)'h+o(h)}
d
d
x
log
a
f
(
x
)
=
d
d
x
ln
f
(
x
)
ln
(
a
)
=
f
′
(
x
)
f
(
x
)
ln
(
a
)
.
{\displaystyle {\frac {d}{dx}}\log _{a}f(x)={\frac {d}{dx}}{\frac {\ln f(x)}{\ln(a)}}={\frac {f'(x)}{f(x)\ln(a)}}.}
d
d
x
sin
x
=
cos
x
{\displaystyle {d \over dx}\sin x=\cos x}
sin
(
x
+
h
)
=
sin
x
+
(
sin
x
)
′
h
+
o
(
h
)
{\displaystyle \sin(x+h)=\sin x+(\sin x)'h+o(h)}
sin
(
x
+
h
)
−
sin
x
=
(
sin
x
)
′
h
+
o
(
h
)
{\displaystyle \sin(x+h)-\sin x=(\sin x)'h+o(h)}
2
sin
h
2
cos
2
x
+
h
2
=
(
sin
x
)
′
h
+
o
(
h
)
{\displaystyle 2\sin {\frac {h}{2}}\cos {\frac {2x+h}{2}}=(\sin x)'h+o(h)}
2
(
h
2
+
o
(
h
)
)
(
cos
x
+
o
(
1
)
)
=
(
sin
x
)
′
h
+
o
(
h
)
{\displaystyle 2({\frac {h}{2}}+o(h))(\cos x+o(1))=(\sin x)'h+o(h)}
(
cos
x
)
h
+
o
(
h
)
=
(
sin
x
)
′
h
+
o
(
h
)
{\displaystyle (\cos x)h+o(h)=(\sin x)'h+o(h)}
cos
x
=
sin
′
x
{\displaystyle \cos x=\sin 'x}
d
d
x
cos
x
=
−
sin
x
{\displaystyle {d \over dx}\cos x=-\sin x}
d
d
x
tg
x
=
sec
2
x
=
1
cos
2
x
=
tg
2
x
+
1
{\displaystyle {d \over dx}\,\operatorname {tg} \,x=\sec ^{2}x={1 \over \cos ^{2}x}=\operatorname {tg} ^{2}x+1}
d
d
x
ctg
x
=
−
cosec
2
x
=
−
1
sin
2
x
{\displaystyle {d \over dx}\,\operatorname {ctg} \,x=-\,\operatorname {cosec} ^{2}\,x=-{1 \over \sin ^{2}x}}
d
d
x
sec
x
=
tg
x
sec
x
{\displaystyle {d \over dx}\sec x=\,\operatorname {tg} \,x\sec x}
d
d
x
cosec
x
=
−
ctg
x
cosec
x
{\displaystyle {d \over dx}\,\operatorname {cosec} \,x=-\,\operatorname {ctg} \,x\,\operatorname {cosec} \,x}
d
d
x
arcsin
x
=
1
1
−
x
2
{\displaystyle {d \over dx}\arcsin x={1 \over {\sqrt {1-x^{2}}}}}
d
d
x
arccos
x
=
−
1
1
−
x
2
{\displaystyle {d \over dx}\arccos x=-{1 \over {\sqrt {1-x^{2}}}}}
d
d
x
arctg
x
=
1
1
+
x
2
{\displaystyle {d \over dx}\,\operatorname {arctg} \,x={1 \over 1+x^{2}}}
d
d
x
arcctg
x
=
−
1
1
+
x
2
{\displaystyle {d \over dx}\,\operatorname {arcctg} \,x=-{1 \over 1+x^{2}}}
d
d
x
arcsec
x
=
1
|
x
|
x
2
−
1
{\displaystyle {d \over dx}\operatorname {arcsec} x={1 \over |x|{\sqrt {x^{2}-1}}}}
d
d
x
arccosec
x
=
−
1
|
x
|
x
2
−
1
{\displaystyle {d \over dx}\,\operatorname {arccosec} \,x=-{1 \over |x|{\sqrt {x^{2}-1}}}}
d
d
x
sh
x
=
ch
x
{\displaystyle {d \over dx}\,\operatorname {sh} \,x=\,\operatorname {ch} \,x}
d
d
x
ch
x
=
sh
x
{\displaystyle {d \over dx}\,\operatorname {ch} \,x=\,\operatorname {sh} \,x}
d
d
x
th
x
=
sech
2
x
=
1
−
th
2
x
{\displaystyle {d \over dx}\,\operatorname {th} \,x=\,\operatorname {sech} ^{2}\,x=1-\operatorname {th} ^{2}\,x}
d
d
x
sech
x
=
−
th
x
sech
x
{\displaystyle {d \over dx}\,\operatorname {sech} \,x=-\operatorname {th} x\,\operatorname {sech} \,x}
d
d
x
cth
x
=
−
csch
2
x
{\displaystyle {d \over dx}\,\operatorname {cth} \,x=-\,\operatorname {csch} ^{2}\,x}
d
d
x
csch
x
=
−
cth
x
csch
x
{\displaystyle {d \over dx}\,\operatorname {csch} \,x=-\,\operatorname {cth} \,x\,\operatorname {csch} \,x}
d
d
x
arsh
x
=
1
x
2
+
1
{\displaystyle {d \over dx}\,\operatorname {arsh} \,x={1 \over {\sqrt {x^{2}+1}}}}
d
d
x
arch
x
=
1
x
2
−
1
{\displaystyle {d \over dx}\,\operatorname {arch} \,x={1 \over {\sqrt {x^{2}-1}}}}
d
d
x
arth
x
=
1
1
−
x
2
{\displaystyle {d \over dx}\,\operatorname {arth} \,x={1 \over 1-x^{2}}}
|
x
|
<
1
{\displaystyle |x|<1}
d
d
x
arsech
x
=
−
1
x
1
−
x
2
{\displaystyle {d \over dx}\,\operatorname {arsech} \,x=-{1 \over x{\sqrt {1-x^{2}}}}}
d
d
x
arcth
x
=
1
1
−
x
2
{\displaystyle {d \over dx}\,\operatorname {arcth} \,x={1 \over 1-x^{2}}}
|
x
|
>
1
{\displaystyle |x|>1}
d
d
x
arcsch
x
=
−
1
|
x
|
1
+
x
2
{\displaystyle {d \over dx}\,\operatorname {arcsch} \,x=-{1 \over |x|{\sqrt {1+x^{2}}}}}
Правила дифференцирования общих функций [ править | править код ]
(
c
f
)
′
=
c
f
′
{\displaystyle \left({cf}\right)'=cf'}
(
f
+
g
)
′
=
f
′
+
g
′
{\displaystyle \left({f+g}\right)'=f'+g'}
(
f
−
g
)
′
=
f
′
−
g
′
{\displaystyle \left({f-g}\right)'=f'-g'}
(
f
g
)
′
=
f
′
g
+
f
g
′
{\displaystyle \left({fg}\right)'=f'g+fg'}
формулы Лейбница )
(
f
g
)
′
=
f
′
g
−
f
g
′
g
2
,
g
≠
0
{\displaystyle \left({f \over g}\right)'={f'g-fg' \over g^{2}},\qquad g\neq 0}
(
f
g
)
′
=
(
e
g
ln
f
)
′
=
f
g
(
f
′
g
f
+
g
′
ln
f
)
,
f
>
0
{\displaystyle (f^{g})'=\left(e^{g\ln f}\right)'=f^{g}\left(f'{g \over f}+g'\ln f\right),\qquad f>0}
(
f
(
g
(
x
)
)
)
′
=
f
′
(
g
(
x
)
)
⋅
g
′
(
x
)
{\displaystyle (f(g(x)))'=f'(g(x))\cdot g'(x)}
Правило дифференцирования сложной функции
f
′
=
(
ln
f
)
′
f
,
f
>
0
{\displaystyle f'=(\ln f)'f,\qquad f>0}
(
f
c
)
′
=
c
(
f
c
−
1
)
f
′
{\displaystyle (f^{c})'=c\left(f^{c-1}\right)f'}
![{\displaystyle {d \over dx}{\sqrt[{n}]{x}}={d \over dx}x^{1 \over n}={1 \over n}x^{1-n \over n}={\frac {1}{n\cdot {\sqrt[{n}]{x^{n-1}}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/08544c34542318779a8cd1718ac6db48fe71e4eb)
