Afgeleide van $$$- \frac{2 x}{x^{2} + 1}$$$
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Uw invoer
Bepaal $$$\frac{d}{dx} \left(- \frac{2 x}{x^{2} + 1}\right)$$$.
Oplossing
Pas de regel van de constante factor $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$ toe met $$$c = -2$$$ en $$$f{\left(x \right)} = \frac{x}{x^{2} + 1}$$$:
$${\color{red}\left(\frac{d}{dx} \left(- \frac{2 x}{x^{2} + 1}\right)\right)} = {\color{red}\left(- 2 \frac{d}{dx} \left(\frac{x}{x^{2} + 1}\right)\right)}$$Pas de quotiëntregel $$$\frac{d}{dx} \left(\frac{f{\left(x \right)}}{g{\left(x \right)}}\right) = \frac{\frac{d}{dx} \left(f{\left(x \right)}\right) g{\left(x \right)} - f{\left(x \right)} \frac{d}{dx} \left(g{\left(x \right)}\right)}{g^{2}{\left(x \right)}}$$$ toe op $$$f{\left(x \right)} = x$$$ en $$$g{\left(x \right)} = x^{2} + 1$$$:
$$- 2 {\color{red}\left(\frac{d}{dx} \left(\frac{x}{x^{2} + 1}\right)\right)} = - 2 {\color{red}\left(\frac{\frac{d}{dx} \left(x\right) \left(x^{2} + 1\right) - x \frac{d}{dx} \left(x^{2} + 1\right)}{\left(x^{2} + 1\right)^{2}}\right)}$$Pas de machtsregel $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ toe met $$$n = 1$$$, met andere woorden, $$$\frac{d}{dx} \left(x\right) = 1$$$:
$$- \frac{2 \left(- x \frac{d}{dx} \left(x^{2} + 1\right) + \left(x^{2} + 1\right) {\color{red}\left(\frac{d}{dx} \left(x\right)\right)}\right)}{\left(x^{2} + 1\right)^{2}} = - \frac{2 \left(- x \frac{d}{dx} \left(x^{2} + 1\right) + \left(x^{2} + 1\right) {\color{red}\left(1\right)}\right)}{\left(x^{2} + 1\right)^{2}}$$De afgeleide van een som/verschil is de som/het verschil van de afgeleiden:
$$- \frac{2 \left(x^{2} - x {\color{red}\left(\frac{d}{dx} \left(x^{2} + 1\right)\right)} + 1\right)}{\left(x^{2} + 1\right)^{2}} = - \frac{2 \left(x^{2} - x {\color{red}\left(\frac{d}{dx} \left(x^{2}\right) + \frac{d}{dx} \left(1\right)\right)} + 1\right)}{\left(x^{2} + 1\right)^{2}}$$De afgeleide van een constante is $$$0$$$:
$$- \frac{2 \left(x^{2} - x \left({\color{red}\left(\frac{d}{dx} \left(1\right)\right)} + \frac{d}{dx} \left(x^{2}\right)\right) + 1\right)}{\left(x^{2} + 1\right)^{2}} = - \frac{2 \left(x^{2} - x \left({\color{red}\left(0\right)} + \frac{d}{dx} \left(x^{2}\right)\right) + 1\right)}{\left(x^{2} + 1\right)^{2}}$$Pas de machtsregel $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ toe met $$$n = 2$$$:
$$- \frac{2 \left(x^{2} - x {\color{red}\left(\frac{d}{dx} \left(x^{2}\right)\right)} + 1\right)}{\left(x^{2} + 1\right)^{2}} = - \frac{2 \left(x^{2} - x {\color{red}\left(2 x\right)} + 1\right)}{\left(x^{2} + 1\right)^{2}}$$Vereenvoudig:
$$- \frac{2 \left(1 - x^{2}\right)}{\left(x^{2} + 1\right)^{2}} = \frac{2 \left(x^{2} - 1\right)}{\left(x^{2} + 1\right)^{2}}$$Dus, $$$\frac{d}{dx} \left(- \frac{2 x}{x^{2} + 1}\right) = \frac{2 \left(x^{2} - 1\right)}{\left(x^{2} + 1\right)^{2}}$$$.
Antwoord
$$$\frac{d}{dx} \left(- \frac{2 x}{x^{2} + 1}\right) = \frac{2 \left(x^{2} - 1\right)}{\left(x^{2} + 1\right)^{2}}$$$A