Kenan kılıçaslan

  • Perdita di attrito
  • Equazione differenziale
  • Risolvi l'equazione
    Hesap Modülleri Matematica

Derivati ​​delle funzioni trigonometriche

1.
$\displaystyle \frac{d}{dx}\left ( \sin u \right ) =\displaystyle \cos u \displaystyle\frac{du}{dx}$
2.
$\displaystyle \frac{d}{dx}\left ( \cos u \right ) =\displaystyle -\sin u \displaystyle\frac{du}{dx}$
3.
$\displaystyle \frac{d}{dx}\left ( \tan u \right ) =\displaystyle \sec^2 u \displaystyle\frac{du}{dx}$
4.
$\displaystyle \frac{d}{dx}\left ( \cot u \right ) =\displaystyle -\csc^2 u \displaystyle\frac{du}{dx}$
5.
$\displaystyle \frac{d}{dx}\left ( \sec u \right ) =\displaystyle \sec u \displaystyle \tan u \displaystyle\frac{du}{dx}$
6.
$\displaystyle \frac{d}{dx}\left ( \csc u \right ) =\displaystyle -\csc u \displaystyle \cot u \displaystyle\frac{du}{dx}$
7.
$\displaystyle \frac{d}{dx}\left ( \sin^{-1} u \right ) =\displaystyle \frac{1}{\displaystyle\sqrt{1-u^2}} \displaystyle\frac{du}{dx}$
8.
$\displaystyle \frac{d}{dx}\left ( \cos^{-1} u \right ) =\displaystyle -\frac{1}{\displaystyle\sqrt{1-u^2}} \displaystyle\frac{du}{dx}$
9.
$\displaystyle \frac{d}{dx}\left ( \tan^{-1} u \right ) =\displaystyle \frac{1}{\displaystyle{1+u^2}} \displaystyle\frac{du}{dx}$
10.
$\displaystyle \frac{d}{dx}\left ( \cot^{-1} u \right ) =\displaystyle -\frac{1}{\displaystyle{1+u^2}} \displaystyle\frac{du}{dx}$
11.
$\displaystyle \frac{d}{dx}\left ( \sec^{-1} u \right ) =\displaystyle \frac{1}{\displaystyle\left | u \right | \sqrt{1+u^2}} \displaystyle\frac{du}{dx}$
12.
$\displaystyle \frac{d}{dx}\left ( \csc^{-1} u \right ) =\displaystyle -\frac{1}{\displaystyle\left | u \right | \sqrt{1+u^2}} \displaystyle\frac{du}{dx}$
beyaz_sayfa_en_alt_oval

Articoli    Documenti    Calcoli    Converti unità    Contatto

Kenan KILIÇASLAN 2012© Copyright.       Designed by Nuit