Pongungaban

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

Montok iso pampos (function) ${\displaystyle f(x)=x^{3}+2x}$, tungab (derivative) dau nopo nga' suaton do ${\displaystyle f'(x)}$ toi ${\displaystyle {\frac {d}{dx}}(f(x))}$. 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:

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

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

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

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

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

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

Haro piipiro tuku^[1] do popouhan pongihiman do tungab di pampos timpou otungab. Id saau, ${\displaystyle x}$ nopo nga' iso osimban (variable), om suai pimato pakayon nopo nga' togom (constant). Nga' montok ii togom ${\displaystyle e}$ kigatang do somok ${\displaystyle e\approx 2.71828...}$

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

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

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

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

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

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

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

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

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

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

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

1. ↑ https://openstax.org/books/calculus-volume-1/pages/3-3-differentiation-rules