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 $f^{\prime }(x)$ 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:

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

Pongomoi

Mamakai do pongomoi dompok $\epsilon$-$\delta$, 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

$L=\underset{\mathrm{\Delta }x\to 0}{\lim }\frac{f(a+\mathrm{\Delta }x)-f(a)}{\mathrm{\Delta }x}$

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

$|\mathrm{\Delta }x|<\delta \text{om}\mathrm{\Delta }x\ne 0$

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

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

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

Tungab pampos tinsait, tintuhun daudau, om tintuhun kiimpou songkukuo

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

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

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

$\frac{d}{dx}({\log }_a(x))=\frac{1}{x\ln (a)}$ $\text{ontok}x,a>0$

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

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

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

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