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.

Tiesiškumas

${\displaystyle \left({cf}\right)'=cf'}$

${\displaystyle \left({f\pm g}\right)'=f'\pm g'}$

Daugybos taisyklė

${\displaystyle \left({fg}\right)'=f'g+fg'}$

Dalybos taisyklė

${\displaystyle \left({f \over g}\right)'={f'g-fg' \over g^{2}},\qquad g\neq 0}$

Eksponentinės funkcijos taisyklė

${\displaystyle (e^{f(x)})'=f'(x)(e^{f(x)})}$

Logaritminės funkcijos taisyklė

${\displaystyle (\ln(f(x)))'={f'(x) \over f(x)}}$

Sudėtinės funkcijos taisyklė

${\displaystyle f'(g(x))=f'(t)g'(x),\,t=g(x)}$

${\displaystyle {d \over dx}c=0}$

${\displaystyle {d \over dx}x=1}$

${\displaystyle {d \over dx}cx=c}$

${\displaystyle {d \over dx}|x|={x \over |x|},\qquad x\neq 0}$

${\displaystyle {d \over dx}x^{c}=cx^{c-1}}$

${\displaystyle {d \over dx}\left({1 \over x}\right)={d \over dx}\left(x^{-1}\right)=-x^{-2}=-{1 \over x^{2}}}$

${\displaystyle {d \over dx}\left({1 \over x^{c}}\right)={d \over dx}\left(x^{-c}\right)=-{c \over x^{c+1}}}$

${\displaystyle {d \over dx}{\sqrt {x}}={d \over dx}x^{1 \over 2}={1 \over 2}x^{-{1 \over 2}}={1 \over 2{\sqrt {x}}},\qquad x>0}$

${\displaystyle {d \over dx}c^{x}=c^{x}\log _{e}c={c^{x}\ln c},\qquad c>0;}$

${\displaystyle {d \over dx}e^{x}=e^{x}\log _{e}e=e^{x};}$

${\displaystyle {d \over dx}e^{-x}=-e^{-x}=\sinh(x)-\cosh(x);}$

${\displaystyle {d \over dx}\log _{c}x={1 \over x\ln c},\qquad c>0,c\neq 1;}$

${\displaystyle {d \over dx}\ln x={1 \over x},\qquad x>0;}$

${\displaystyle {d \over dx}\ln |x|={1 \over x};}$

${\displaystyle {d \over dx}x^{x}=x^{x}(1+\ln x).}$

${\displaystyle {d \over dx}\sin x=\cos x}$

${\displaystyle {d \over dx}\cos x=-\sin x}$

${\displaystyle {d \over dx}\tan x=\sec ^{2}x={1 \over \cos ^{2}x}}$

${\displaystyle {d \over dx}\sec x=\tan x\sec x={\frac {\sin x}{\cos ^{2}x}}}$

${\displaystyle {d \over dx}\csc x=-\csc x\cot x=-{\frac {\cos x}{\sin ^{2}x}}}$

${\displaystyle {d \over dx}\cot x=-\csc ^{2}x={-1 \over \sin ^{2}x}}$

${\displaystyle {d \over dx}\arcsin x={1 \over {\sqrt {1-x^{2}}}}}$

${\displaystyle {d \over dx}\arccos x={-1 \over {\sqrt {1-x^{2}}}}}$

${\displaystyle {d \over dx}\arctan x={1 \over 1+x^{2}}}$

${\displaystyle {d \over dx}\operatorname {arcsec} x={1 \over |x|{\sqrt {x^{2}-1}}}}$

${\displaystyle {d \over dx}\operatorname {arccsc} x={-1 \over |x|{\sqrt {x^{2}-1}}}}$

${\displaystyle {d \over dx}\operatorname {arccot} x={-1 \over 1+x^{2}}}$

${\displaystyle {d \over dx}\sinh x=\cosh x={\frac {e^{x}+e^{-x}}{2}}}$

${\displaystyle {d \over dx}\cosh x=\sinh x={\frac {e^{x}-e^{-x}}{2}}}$

${\displaystyle {d \over dx}\tanh x=\operatorname {sech} ^{2}\,x}$

${\displaystyle {d \over dx}\,\operatorname {sech} \,x=-\tanh x\,\operatorname {sech} \,x}$

${\displaystyle {d \over dx}\,\operatorname {coth} \,x=-\,\operatorname {csch} ^{2}\,x}$

${\displaystyle {d \over dx}\,\operatorname {csch} \,x=-\,\operatorname {coth} \,x\,\operatorname {csch} \,x}$

${\displaystyle {d \over dx}\,\operatorname {arcsinh} \,x={1 \over {\sqrt {x^{2}+1}}}}$

${\displaystyle {d \over dx}\,\operatorname {arccosh} \,x={-1 \over {\sqrt {x^{2}-1}}}}$

${\displaystyle {d \over dx}\,\operatorname {arctanh} \,x={1 \over 1-x^{2}}}$

${\displaystyle {d \over dx}\,\operatorname {arcsech} \,x={-1 \over x{\sqrt {1-x^{2}}}}}$

${\displaystyle {d \over dx}\,\operatorname {arccoth} \,x={1 \over 1-x^{2}}}$

${\displaystyle {d \over dx}\,\operatorname {arccsch} \,x={-1 \over |x|{\sqrt {1+x^{2}}}}}$

${\displaystyle {d \over dx}(f^{-1}(x))={\frac {1}{f'(f^{-1}(x))}}}$, bet kuriai diferencijuojamai realaus argumento funkcijai f su realiomis vertėmis, kada surasta kompozicija ir inversija egzistuoja.