Вычисление производной — операция в дифференциальном исчислении . Эта статья содержит список формул для нахождения производных от некоторых функций.
В этих формулах f {\displaystyle f} и g {\displaystyle g} — произвольные дифференцируемые функции вещественной переменной , а c {\displaystyle c} — вещественная константа. Этих формул достаточно для дифференцирования любой 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},} когда x c {\displaystyle x^{c}} и c x c − 1 {\displaystyle cx^{c-1}} определены, 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}}}} , то пусть g ( x ) = x 2 , h ( x ) = x {\displaystyle g(x)=x^{2},\quad h(x)={\sqrt {x}}} и 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}}} , при | 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}}} , при | 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}}}}} ( 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'} (частный случай 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'}