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Derivatan av $$$x^{x}$$$
Relaterad kalkylator: Derivata-beräknare
Din inmatning
Bestäm $$$\frac{d}{dx} \left(x^{x}\right)$$$.
Lösning
Låt $$$H{\left(x \right)} = x^{x}$$$.
Ta logaritmen av båda sidorna: $$$\ln\left(H{\left(x \right)}\right) = \ln\left(x^{x}\right)$$$.
Skriv om högerledet med hjälp av logaritmlagarna: $$$\ln\left(H{\left(x \right)}\right) = x \ln\left(x\right)$$$.
Derivera båda leden i ekvationen var för sig: $$$\frac{d}{dx} \left(\ln\left(H{\left(x \right)}\right)\right) = \frac{d}{dx} \left(x \ln\left(x\right)\right)$$$.
Derivera ekvationens vänsterled.
Funktionen $$$\ln\left(H{\left(x \right)}\right)$$$ är sammansättningen $$$f{\left(g{\left(x \right)} \right)}$$$ av två funktioner $$$f{\left(u \right)} = \ln\left(u\right)$$$ och $$$g{\left(x \right)} = H{\left(x \right)}$$$.
Tillämpa kedjeregeln $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$:
$${\color{red}\left(\frac{d}{dx} \left(\ln\left(H{\left(x \right)}\right)\right)\right)} = {\color{red}\left(\frac{d}{du} \left(\ln\left(u\right)\right) \frac{d}{dx} \left(H{\left(x \right)}\right)\right)}$$Derivatan av den naturliga logaritmen är $$$\frac{d}{du} \left(\ln\left(u\right)\right) = \frac{1}{u}$$$:
$${\color{red}\left(\frac{d}{du} \left(\ln\left(u\right)\right)\right)} \frac{d}{dx} \left(H{\left(x \right)}\right) = {\color{red}\left(\frac{1}{u}\right)} \frac{d}{dx} \left(H{\left(x \right)}\right)$$Återgå till den ursprungliga variabeln:
$$\frac{\frac{d}{dx} \left(H{\left(x \right)}\right)}{{\color{red}\left(u\right)}} = \frac{\frac{d}{dx} \left(H{\left(x \right)}\right)}{{\color{red}\left(H{\left(x \right)}\right)}}$$Alltså, $$$\frac{d}{dx} \left(\ln\left(H{\left(x \right)}\right)\right) = \frac{\frac{d}{dx} \left(H{\left(x \right)}\right)}{H{\left(x \right)}}$$$.
Derivera ekvationens högerled.
Tillämpa produktregeln $$$\frac{d}{dx} \left(f{\left(x \right)} g{\left(x \right)}\right) = \frac{d}{dx} \left(f{\left(x \right)}\right) g{\left(x \right)} + f{\left(x \right)} \frac{d}{dx} \left(g{\left(x \right)}\right)$$$ med $$$f{\left(x \right)} = x$$$ och $$$g{\left(x \right)} = \ln\left(x\right)$$$:
$${\color{red}\left(\frac{d}{dx} \left(x \ln\left(x\right)\right)\right)} = {\color{red}\left(\frac{d}{dx} \left(x\right) \ln\left(x\right) + x \frac{d}{dx} \left(\ln\left(x\right)\right)\right)}$$Derivatan av den naturliga logaritmen är $$$\frac{d}{dx} \left(\ln\left(x\right)\right) = \frac{1}{x}$$$:
$$x {\color{red}\left(\frac{d}{dx} \left(\ln\left(x\right)\right)\right)} + \ln\left(x\right) \frac{d}{dx} \left(x\right) = x {\color{red}\left(\frac{1}{x}\right)} + \ln\left(x\right) \frac{d}{dx} \left(x\right)$$Tillämpa potensregeln $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ med $$$n = 1$$$, det vill säga $$$\frac{d}{dx} \left(x\right) = 1$$$:
$$\ln\left(x\right) {\color{red}\left(\frac{d}{dx} \left(x\right)\right)} + 1 = \ln\left(x\right) {\color{red}\left(1\right)} + 1$$Alltså, $$$\frac{d}{dx} \left(x \ln\left(x\right)\right) = \ln\left(x\right) + 1$$$.
Således, $$$\frac{\frac{d}{dx} \left(H{\left(x \right)}\right)}{H{\left(x \right)}} = \ln\left(x\right) + 1$$$.
Således, $$$\frac{d}{dx} \left(H{\left(x \right)}\right) = \left(\ln\left(x\right) + 1\right) H{\left(x \right)} = x^{x} \left(\ln\left(x\right) + 1\right)$$$.
Svar
$$$\frac{d}{dx} \left(x^{x}\right) = x^{x} \left(\ln\left(x\right) + 1\right)$$$A