määritelmä

Lause

$Dx^n=nx^{n-1}{,}\ kun\ x\ne0\ ja\ n\in\mathbb{Z}$

Muista!

$x^{-k}=\frac{1}{x^k}$

Esimerkki. Derivoi

$\frac{3}{x^3}{,}\ x\ne0$
$D\ \frac{3}{x^3}=D\ 3\cdot\frac{1}{x^3}=3x^{-3}=3\left(-3\right)x^{-3-1}=-9x^{-4}=-\frac{9}{x^4}$

$\frac{x^4-2}{x^4}{,}x\ne0\ \ D\left(\frac{x^4}{x^4}-\frac{2}{x^4}\right)=D\left(1-\frac{2}{x^4}\right)$
$=D\left(1-2x^{-4}\right)=8x^{-5}=\frac{8}{x^{-5}}$

Lause

Olkoon f ja g derivoituvia. Tällöin

$D\left(f\left(x\right)g\left(x\right)\right)=f'\left(x\right)g\left(x\right)+f\left(x\right)g'\left(x\right)$

$D\left(\frac{f\left(x\right)}{g\left(x\right)}\right)=\frac{f'\left(x\right)g\left(x\right)-f\left(x\right)g'\left(x\right)}{\left(g\left(x\right)\right)^2}{,}\ kun\ g\left(x\right)\ne0$

Esim

$D\left(2x\left(x^2+1\right)\right)=$

nyt

$f\left(x\right)=2x{,}\ f\left(x\right)=2\ ja\ g\left(x\right)=x^2+1{,}\ g'\left(x\right)=2x$

siis

$D\left(2x\left(x^2+1\right)\right)=2x\cdot2x+2\left(x^2+1\right)=4x^2+2x^2+2=6x^2+2$

toinen tapa

$D\left(2x\left(x^2+1\right)\right)=2x^3+2x=6x^2+2$

$D\ \frac{x^4-2}{x^4}$

nyt

$f\left(x\right)=x^4-2=f'\left(x\right)=4x^3$
$g\left(x\right)=x^4{,}\ g'\left(x\right)=4x^3$

siis

$D\ \frac{x^4-2}{x^4}=\frac{4x^3\cdot x^4-\left(x^4-2\right)4x^3}{\left(x^4\right)^2}=\frac{4x^7-4x^7+8x^3}{x^8}=\frac{8x^3}{x^8}=\frac{8}{x^5}{,}\ x\ne0$