Pongungaban

Pongungaban (Boros Inggilis: differentiation) nopo nga' iso katandaan id raan gana' pongintaban do Sidsimban (calculus). Katandaan diti sambalik do Ponompuuvan (integration).

Poomitanan

Montok iso pampos (function) $f(x)=x^{3}+2x$, tungab (derivative) dau nopo nga' suaton do $f'(x)$ toi ${\frac {d}{dx}}(f(x))$. It koiso nopo nga' roitan do wayagait di Newton, om it koduo wayagait di Leibniz.

Momoguno do tuku sundu (power rule), ii tungab nopo nga:

$f'(x)=3x^{2}+2$

Pongomoi

Mamakai do pongomoi dompok $\varepsilon$-$\delta$, montok iso pampos kigatang totunud (real-valued function), karait tokou id sogom $a$ suang raung dau, om haro id suang raung dau tokuri sonsopo poingukab (open interval) poingion di sogom $a$, om hino no i dompok $L$ poingkaa

$L=\lim _{\Delta x\rightarrow 0}{\frac {f(a+\Delta x)-f(a)}{\Delta x}}$

Montok tikid tintob totunud $\varepsilon$ haro nogi iso tintob totunud danan (positive real number), $\delta$ moi do montok tikid $\Delta x$ om poingkaa:

$|\Delta x|<\delta \quad {\textrm {om}}\quad \Delta x\neq 0$

Sumusuut oworun no daa $f(a+\Delta x)$ om $\left|L-{\frac {f(a+\Delta x)-f(a)}{\Delta x}}\right|<\varepsilon$.

Tuku pongungaban

Haro piipiro tuku do popouhan pongihiman do tungab di pampos timpou otungab. Id saau, $x$ nopo nga' iso osimban (variable), om suai pimato pakayon nopo nga' togom (constant). Nga' montok ii togom $e$ kigatang do somok $e\approx 2.71828...$

Tuku sundu

${\frac {d}{dx}}(x^{a})=ax^{a-1}$

Tungab pampos tinsait, tintuhun daudau, om tintuhun kiimpou songkukuo

${\frac {d}{dx}}(e^{x})=e^{x}$

${\frac {d}{dx}}(a^{x})=a^{x}\ln {a}$ $\quad {\textrm {ontok}}\quad a>0$

${\frac {d}{dx}}(\ln {(x)})={\frac {1}{x}}$ $\quad {\textrm {ontok}}\quad x>0$

${\frac {d}{dx}}(\log _{a}{(x)})={\frac {1}{x\ln {(a)}}}$ $\quad {\textrm {ontok}}\quad x,a>0$

Tungab pampos tolusunduk

${\frac {d}{dx}}(\sin(x))=\cos(x)$

${\frac {d}{dx}}(\cos(x))=-\sin(x)$

${\frac {d}{dx}}(\tan(x))=\sec ^{2}(x)={\frac {1}{1+\cos ^{2}(x)}}=1+\tan ^{2}(x)$

Tungab pampos bolik tolusunduk

${\frac {d}{dx}}(\arcsin(x))={\frac {1}{\sqrt {1-x^{2}}}}$ $\quad {\textrm {ontok}}\quad x\in (-1,1)$

${\frac {d}{dx}}(\arccos(x))=-{\frac {1}{\sqrt {1-x^{2}}}}$ $\quad {\textrm {ontok}}\quad x\in (-1,1)$

${\frac {d}{dx}}(\arctan(x))={\frac {1}{1+x^{2}}}$