Pravila za odvajanje

Irina Cristea; Hashem Bordbar; Alessandro Linzi

15 Pravila za odvajanje

Izrek 15.1 [9, 10]: Če sta funkciji $f$ in $g$ odvedljivi v točki $x$ , je v tej točki odvedljiva tudi funkcija:

$f \pm g$ in velja $(f \pm g)' (x) = f'(x) \pm g'(x)$ ,
$k \cdot f$ in velja $(k \cdot f)' (x) = k \cdot f'(x)$ ,
$f \cdot g$ in velja $(f \cdot g)' (x) = f'(x) \cdot g(x) + f(x) \cdot g'(x)$ ,
$\dfrac{f}{g}$ , ko je $g(x) \neq 0$ , in velja $\left(\dfrac{f}{g}\right)' (x) = \dfrac{f'(x) \cdot g(x) - f(x) \cdot g'(x)}{g^2(x)}$ ,
$g\circ f$ in velja $(g\circ f)' (x) = f'(x) \cdot g'(f(x))$ .

Tabela odvodov elementarnih funkcij:

$k' = 0$ (ker je $k$ poljubna konstanta),
$(x^n)' = n x^{n-1}$ , $x' = 1, (x^{-1})'= -x^{-2} = -\frac{1}{x^2}$ ,
$(e^x)' = e^{x}$ ,
$(a^x)' = a^{x} \cdot \ln a$ ,
$(\ln x)' = \dfrac{1}{x}$ ,
$(\log_a x)' = \dfrac{1}{x \ln a}$ ,
$(\sin x)' = \cos x$ ,
$(\cos x)' = - \sin x$ ,
$(\tan x)' = \dfrac{1}{\cos^2 x}$ ,
$(\cot x)' =- \dfrac{1}{\sin^2 x}$ ,
$(\arcsin x)' = \dfrac{1}{\sqrt{1-x^2}}$ ,
$(\arccos x)' = -\dfrac{1}{\sqrt{1-x^2}}$ ,
$(\arctan x)' = \dfrac{1}{1+x^2}$ .

Primer 15.1: Po formulah tako dobimo:

$(x^{10})' = 10x^9, (x^{-5})' = -5x^{-6}, (x^{-\frac{1}{3}})' = -\dfrac{1}{3}x^{-\frac{4}{3}},$
$(\sqrt[3]{x^2})' = (x^{\frac{2}{3}})' = \dfrac{2}{3} x^{-\frac{1}{3}},$
$(\sqrt{x^{-3}})' = (x^{-\frac{3}{2}})' = -\dfrac{3}{2} x^{-\frac{5}{2}}.$

Primer 15.2: Izračunajmo odvod funkcije.

$f(x) = (x^2+x+1)^9.$
$f'(x) = 9 \cdot (x^2+ x+1)^8 \cdot (x^2+x+1)'$
$= 9 \cdot (x^2+ x+1)^8 \cdot (2x+1).$
$f(x) = (x^{-2}+x^{-1}+4)^{\frac{2}{5}}.$
$f'(x) = \frac{2}{5} \cdot (x^{-2}+x^{-1}+4)^{-\frac{3}{5}} \cdot (x^{-2}+x^{-1}+4)'$
$= \frac{2}{5} \cdot (x^{-2}+x^{-1}+4)^{-\frac{3}{5}} \cdot (-2x^{-3}-x^{-2}).$
$f(x) = \sqrt{x^2 - 3 + \dfrac{1}{x}}.$
$f'(x) = ((x^2 - 3 + \dfrac{1}{x})^{\frac{1}{2}})'= \dfrac{1}{2} \cdot (x^2 - 3 + \dfrac{1}{x})^{-\frac{1}{2}} \cdot (x^2 - 3 + \dfrac{1}{x})'$
$=\dfrac{1}{2} \cdot (x^2 - 3 + \dfrac{1}{x})^{-\frac{1}{2}} \cdot (2x - \dfrac{1}{x^2}).$
$f(x) = e^{x^2}.$
$f'(x)=(e^{x^2})' = e^{x^2}\cdot (x^2)' =e^{x^2}\cdot (2x).$
$f(x) = e^{(x^2 + 3x - 4)}.$
$f'(x) = (e^{(x^2 + 3x - 4)})' = e^{(x^2 + 3x - 4)}\cdot (x^2 + 3x - 4)'$
$=e^{(x^2 + 3x - 4)}\cdot (2x+3).$
$f(x) = e^{\sqrt{x}}.$
$f'(x) = (e^{\sqrt{x}})' = (\sqrt{x})' \cdot e^{\sqrt{x}} = \dfrac{1}{2\sqrt{x}}\cdot e^{\sqrt{x}}.$
$f(x) = e^{\sqrt[3]{x^5}}.$
$f'(x) = (e^{\sqrt[3]{x^5}})' = e^{\sqrt[3]{x^5}} \cdot (\sqrt[3]{x^5})' = e^{\sqrt[3]{x^5}}\cdot ({x^\frac{5}{3}})' = e^{\sqrt[3]{x^5}}\cdot \dfrac{5}{3} \cdot x^{\frac{2}{3}}.$
$f(x) = (x+3) \cdot e^{2x}.$
$f'(x) = (x+3)' \cdot e^{2x} + (x+3) \cdot (e^{2x})' = e^{2x} + (x+3) \cdot e^{2x} \cdot (2x)'$
$= e^{2x} + (x+3) \cdot 2 \cdot e^{2x}.$
$f(x) = \dfrac{1+\ln x}{1 - \ln (2x)}.$
$\begin{align*} f'(x) &= \frac{(1+ \ln x)' \cdot (1 - \ln (2x)) - (1+\ln x) \cdot (1 - \ln(2x))'}{(1-\ln(2x))^2}\\ &= \frac{\dfrac{1}{x} \cdot (1 - \ln (2x)) - (1+\ln x) \cdot (-\dfrac{1}{2x} \cdot 2)}{(1-\ln(2x))^2}\\ &=\frac{(1 - \ln (2x))+(1+\ln x) }{x\cdot (1-\ln (2x))^2}. \end{align*}$
$f(x) = \ln (\dfrac{x+1}{x-1}).$
$\begin{align*} f'(x)& = \frac{1}{\dfrac{x+1}{x-1}}\cdot \left(\frac{x+1}{x-1}\right)'\\ &=\frac{x-1}{x+1}\cdot \frac{(x-1)\cdot (x+1)'-(x+1)\cdot (x-1)'}{(x-1)^2}\\ &=\frac{x-1-(x+1)}{(x+1)(x-1)}=-\frac{2}{x^2-1}. \end{align*}$
$f(x)= \sqrt{\ln (2x)}.$
$\begin{align*} f'(x)& =(\sqrt{\ln (2x)})' = \left[(\ln (2x))^{\frac{1}{2}}\right]' = \frac{1}{2} (\ln (2x))^{-\frac{1}{2}} \cdot (\ln (2x))'\\ &= \frac{1}{2} (\ln (2x))^{-\frac{1}{2}} \cdot \frac{1}{2x}\cdot (2x)'= \frac{1}{2\sqrt{\ln (2x)}}\cdot \frac{1}{2x}\cdot 2\\ &=\frac{1}{2x\sqrt{\ln (2x)}}. \end{align*}$
$f(x) = \ln (\sin x).$
$f'(x) =\left[\ln (\sin x)\right]' = \dfrac{1}{\sin x}\cdot (\sin x)'= \dfrac{1}{\sin x}\cdot \cos x = \cot x.$
$f(x) = \sin {\sqrt{x}}$
$f'(x) = (\sin {\sqrt{x}})'=\cos {\sqrt{x}}\cdot (\sqrt{x})'= \cos {\sqrt{x}} \cdot \frac{1}{2} x^{-\frac{1}{2}}= \cos {\sqrt{x}} \cdot \dfrac{1}{2\sqrt{x}}.$

Primer 15.3: Izračunajmo odvod funkcije $f(x) = x^x$ .

Najprej dano funkcijo napišemo kot eksponentno funkcijo

$f(x) = x^x= e^{\ln (x^x)} = e^{x \cdot \ln x}$

in potem izračunamo

$\begin{align*} f'(x) &= (e^{x \cdot \ln x})'= e^{x \cdot \ln x} \cdot (x \cdot \ln x)'\\ &= e^{x \cdot \ln x} \cdot (x' \cdot \ln x + x \cdot (\ln x)')\\ &= e^{x \cdot \ln x} \cdot (\ln x +1). \end{align*}$

15 Pravila za odvajanje

License

Share This Book