5.1 Rationaalifunktion derivointi

508

a)
$D\ \frac{1}{x^2}=$
$f\left(x\right)=1{,}\ f'\left(x\right)=0\ ja\ g\left(x\right)=x^2{,}\ g'\left(x\right)=2x$

$\frac{0\left(x^2\right)-2x}{x^4}=\frac{-2x}{x^4}=\frac{-2}{x^3}$

$D\ \frac{1}{x^2}=1\cdot x^{-2}=-2x^{-3}=\frac{-2}{x^{-3}}$

507

a)
$D\left(\frac{2}{2x+4}\right)=$
$f\left(x\right)=2{,}\ f'\left(x\right)=0\ ja\ g\left(x\right)=2x+4{,}\ g'\left(x\right)=2$
$\frac{0\left(2x+4\right)-2\cdot2}{\left(2x+4\right)^2}=-\frac{4}{\left(2x+4\right)^2}=-\frac{1}{\left(x+1\right)^2}$
b)
$D\left(\frac{2x+4}{2}\right)=$
$f\left(x\right)=2x+4{,}\ f'\left(x\right)=2\ ja\ g\left(x\right)=2{,}\ g'\left(x\right)=0$
$\frac{2\cdot2-0\left(2x+4\right)}{2^2}=\frac{4}{4}=1$

504

$D\left(\frac{3x}{x+1}\right)=$

$f\left(x\right)=3x{,}\ f'\left(x\right)=3\ ja\ g\left(x\right)=x+1{,}\ g'\left(x\right)=1$
$\frac{3\left(x+1\right)-3x\cdot1}{\left(x+1\right)^2}=\frac{3x+3-3x}{\left(x+1\right)^2}=\frac{3}{\left(x+1\right)^2}$

$D\left(\frac{x^2}{2x+1}\right)$
$f\left(x\right)=x^2{,}\ f'\left(x\right)=2x\ ja\ g\left(x\right)=2x+1{,}\ g'\left(x\right)=2$
$\frac{2x\left(2x+1\right)-2x^2}{\left(2x+1\right)^2}=\frac{4x^2+2x-2x^2}{\left(2x+1\right)^2}=\frac{2x^2+2x}{\left(2x+1\right)^2}$

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

503

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

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

$D\left(f\left(x\right)\left(g\left(x\right)\right)\right)=f\left(x\right)g'\left(x\right)+f'\left(x\right)g\left(x\right)=\left(x^2+1\right)\cdot2x+2x\left(x^2-1\right)$
$=2x^3+2x+2x^3-2x=4x^3$

$\left(x^2+1\right)\left(x^2-1\right)=x^4-x^2+x^2-1=x^4-1=4x^3$

501

$D\left(\frac{1}{x^2}\right)=x^{-2}$

$D\left(\frac{4}{x^3}\right)=4\cdot\frac{1}{x^3}=4x^{-3}=12x^{-4}=\frac{12}{x^4}$

$D\left(-\frac{6}{x^7}\right)=-6x^{-7}=-42x^{-8}=-\frac{42}{x^{-8}}$

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

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