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Funktion $$$5 x^{x}$$$ derivaatta
Aiheeseen liittyvät laskurit: Logaritmisen derivoinnin laskin, Vaiheittainen implisiittisen derivoinnin laskin
Syötteesi
Määritä $$$\frac{d}{dx} \left(5 x^{x}\right)$$$.
Ratkaisu
Sovella vakion kerroinsääntöä $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$ käyttäen $$$c = 5$$$ ja $$$f{\left(x \right)} = x^{x}$$$:
$${\color{red}\left(\frac{d}{dx} \left(5 x^{x}\right)\right)} = {\color{red}\left(5 \frac{d}{dx} \left(x^{x}\right)\right)}$$Käytä kaavaa $$$f^{g{\left(x \right)}}{\left(x \right)} = e^{g{\left(x \right)} \ln\left(f{\left(x \right)}\right)}$$$ $$$f{\left(x \right)} = x$$$:n ja $$$g{\left(x \right)} = x$$$:n kanssa kirjoittaaksesi monimutkaisen lausekkeen uudelleen:
$$5 {\color{red}\left(\frac{d}{dx} \left(x^{x}\right)\right)} = 5 {\color{red}\left(\frac{d}{dx} \left(e^{x \ln\left(x\right)}\right)\right)}$$Funktio $$$e^{x \ln\left(x\right)}$$$ on kahden funktion $$$f{\left(u \right)} = e^{u}$$$ ja $$$g{\left(x \right)} = x \ln\left(x\right)$$$ yhdistelmä $$$f{\left(g{\left(x \right)} \right)}$$$.
Sovella ketjusääntöä $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$:
$$5 {\color{red}\left(\frac{d}{dx} \left(e^{x \ln\left(x\right)}\right)\right)} = 5 {\color{red}\left(\frac{d}{du} \left(e^{u}\right) \frac{d}{dx} \left(x \ln\left(x\right)\right)\right)}$$Eksponenttifunktion derivaatta on $$$\frac{d}{du} \left(e^{u}\right) = e^{u}$$$:
$$5 {\color{red}\left(\frac{d}{du} \left(e^{u}\right)\right)} \frac{d}{dx} \left(x \ln\left(x\right)\right) = 5 {\color{red}\left(e^{u}\right)} \frac{d}{dx} \left(x \ln\left(x\right)\right)$$Palaa alkuperäiseen muuttujaan:
$$5 e^{{\color{red}\left(u\right)}} \frac{d}{dx} \left(x \ln\left(x\right)\right) = 5 e^{{\color{red}\left(x \ln\left(x\right)\right)}} \frac{d}{dx} \left(x \ln\left(x\right)\right) = 5 x^{x} \frac{d}{dx} \left(x \ln\left(x\right)\right)$$Sovella tulon derivointisääntöä $$$\frac{d}{dx} \left(f{\left(x \right)} g{\left(x \right)}\right) = \frac{d}{dx} \left(f{\left(x \right)}\right) g{\left(x \right)} + f{\left(x \right)} \frac{d}{dx} \left(g{\left(x \right)}\right)$$$ funktioille $$$f{\left(x \right)} = x$$$ ja $$$g{\left(x \right)} = \ln\left(x\right)$$$:
$$5 x^{x} {\color{red}\left(\frac{d}{dx} \left(x \ln\left(x\right)\right)\right)} = 5 x^{x} {\color{red}\left(\frac{d}{dx} \left(x\right) \ln\left(x\right) + x \frac{d}{dx} \left(\ln\left(x\right)\right)\right)}$$Luonnollisen logaritmin derivaatta on $$$\frac{d}{dx} \left(\ln\left(x\right)\right) = \frac{1}{x}$$$:
$$5 x^{x} \left(x {\color{red}\left(\frac{d}{dx} \left(\ln\left(x\right)\right)\right)} + \ln\left(x\right) \frac{d}{dx} \left(x\right)\right) = 5 x^{x} \left(x {\color{red}\left(\frac{1}{x}\right)} + \ln\left(x\right) \frac{d}{dx} \left(x\right)\right)$$Sovella potenssisääntöä $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ käyttäen $$$n = 1$$$, toisin sanoen, $$$\frac{d}{dx} \left(x\right) = 1$$$:
$$5 x^{x} \left(\ln\left(x\right) {\color{red}\left(\frac{d}{dx} \left(x\right)\right)} + 1\right) = 5 x^{x} \left(\ln\left(x\right) {\color{red}\left(1\right)} + 1\right)$$Näin ollen, $$$\frac{d}{dx} \left(5 x^{x}\right) = 5 x^{x} \left(\ln\left(x\right) + 1\right)$$$.
Vastaus
$$$\frac{d}{dx} \left(5 x^{x}\right) = 5 x^{x} \left(\ln\left(x\right) + 1\right)$$$A