Diferencialiniame skaičiavime pagrindinis tikslas yra surasti išvestinę. Šiame sąraše pateikiama daugybės matematinių funkcijų išvestinės. Toliau, f ir g yra diferencijuojamos realaus argumento funkcijos, ir c yra realusis skaičius. Šių formulių pakanka bet kokios elementarios funkcijos išvestinėms surasti.
redaguoti Tiesiškumas ( c f ) ′ = c f ′ {\displaystyle \left({cf}\right)'=cf'} ( f ± g ) ′ = f ′ ± g ′ {\displaystyle \left({f\pm g}\right)'=f'\pm g'} Daugybos taisyklė ( f g ) ′ = f ′ g + f g ′ {\displaystyle \left({fg}\right)'=f'g+fg'} Dalybos taisyklė ( 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} Eksponentinės funkcijos taisyklė ( e f ( x ) ) ′ = f ′ ( x ) ( e f ( x ) ) {\displaystyle (e^{f(x)})'=f'(x)(e^{f(x)})} Logaritminės funkcijos taisyklė ( ln ( f ( x ) ) ) ′ = f ′ ( x ) f ( x ) {\displaystyle (\ln(f(x)))'={f'(x) \over f(x)}} Sudėtinės funkcijos taisyklė f ′ ( g ( x ) ) = f ′ ( t ) g ′ ( x ) , t = g ( x ) {\displaystyle f'(g(x))=f'(t)g'(x),\,t=g(x)}
redaguoti 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} d d x | x | = x | x | , x ≠ 0 {\displaystyle {d \over dx}|x|={x \over |x|},\qquad x\neq 0} d d x x c = c x c − 1 {\displaystyle {d \over dx}x^{c}=cx^{c-1}} 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}
redaguoti d d x c x = c x log e c = c x ln c , c > 0 ; {\displaystyle {d \over dx}c^{x}=c^{x}\log _{e}c={c^{x}\ln c},\qquad c>0;} d d x e x = e x log e e = e x ; {\displaystyle {d \over dx}e^{x}=e^{x}\log _{e}e=e^{x};} d d x e − x = − e − x = sinh ( x ) − cosh ( x ) ; {\displaystyle {d \over dx}e^{-x}=-e^{-x}=\sinh(x)-\cosh(x);} d d x log c x = 1 x ln c , c > 0 , c ≠ 1 ; {\displaystyle {d \over dx}\log _{c}x={1 \over x\ln c},\qquad c>0,c\neq 1;} d d x ln x = 1 x , x > 0 ; {\displaystyle {d \over dx}\ln x={1 \over x},\qquad x>0;} d d x ln | x | = 1 x ; {\displaystyle {d \over dx}\ln |x|={1 \over x};} d d x x x = x x ( 1 + ln x ) . {\displaystyle {d \over dx}x^{x}=x^{x}(1+\ln x).}
redaguoti d d x sin x = cos x {\displaystyle {d \over dx}\sin x=\cos x} d d x cos x = − sin x {\displaystyle {d \over dx}\cos x=-\sin x} d d x tan x = sec 2 x = 1 cos 2 x {\displaystyle {d \over dx}\tan x=\sec ^{2}x={1 \over \cos ^{2}x}} d d x sec x = tan x sec x = sin x cos 2 x {\displaystyle {d \over dx}\sec x=\tan x\sec x={\frac {\sin x}{\cos ^{2}x}}} d d x csc x = − csc x cot x = − cos x sin 2 x {\displaystyle {d \over dx}\csc x=-\csc x\cot x=-{\frac {\cos x}{\sin ^{2}x}}} d d x cot x = − csc 2 x = − 1 sin 2 x {\displaystyle {d \over dx}\cot x=-\csc ^{2}x={-1 \over \sin ^{2}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 arctan x = 1 1 + x 2 {\displaystyle {d \over dx}\arctan 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 arccsc x = − 1 | x | x 2 − 1 {\displaystyle {d \over dx}\operatorname {arccsc} x={-1 \over |x|{\sqrt {x^{2}-1}}}} d d x arccot x = − 1 1 + x 2 {\displaystyle {d \over dx}\operatorname {arccot} x={-1 \over 1+x^{2}}}
redaguoti d d x sinh x = cosh x = e x + e − x 2 {\displaystyle {d \over dx}\sinh x=\cosh x={\frac {e^{x}+e^{-x}}{2}}} d d x cosh x = sinh x = e x − e − x 2 {\displaystyle {d \over dx}\cosh x=\sinh x={\frac {e^{x}-e^{-x}}{2}}} d d x tanh x = sech 2 x {\displaystyle {d \over dx}\tanh x=\operatorname {sech} ^{2}\,x} d d x sech x = − tanh x sech x {\displaystyle {d \over dx}\,\operatorname {sech} \,x=-\tanh x\,\operatorname {sech} \,x} d d x coth x = − csch 2 x {\displaystyle {d \over dx}\,\operatorname {coth} \,x=-\,\operatorname {csch} ^{2}\,x} d d x csch x = − coth x csch x {\displaystyle {d \over dx}\,\operatorname {csch} \,x=-\,\operatorname {coth} \,x\,\operatorname {csch} \,x} d d x arcsinh x = 1 x 2 + 1 {\displaystyle {d \over dx}\,\operatorname {arcsinh} \,x={1 \over {\sqrt {x^{2}+1}}}} d d x arccosh x = − 1 x 2 − 1 {\displaystyle {d \over dx}\,\operatorname {arccosh} \,x={-1 \over {\sqrt {x^{2}-1}}}} d d x arctanh x = 1 1 − x 2 {\displaystyle {d \over dx}\,\operatorname {arctanh} \,x={1 \over 1-x^{2}}} d d x arcsech x = − 1 x 1 − x 2 {\displaystyle {d \over dx}\,\operatorname {arcsech} \,x={-1 \over x{\sqrt {1-x^{2}}}}} d d x arccoth x = 1 1 − x 2 {\displaystyle {d \over dx}\,\operatorname {arccoth} \,x={1 \over 1-x^{2}}} d d x arccsch x = − 1 | x | 1 + x 2 {\displaystyle {d \over dx}\,\operatorname {arccsch} \,x={-1 \over |x|{\sqrt {1+x^{2}}}}}
redaguoti d d x ( f − 1 ( x ) ) = 1 f ′ ( f − 1 ( x ) ) {\displaystyle {d \over dx}(f^{-1}(x))={\frac {1}{f'(f^{-1}(x))}}} f su realiomis vertėmis, kada surasta kompozicija ir inversija egzistuoja.