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Funktion $$$\operatorname{atan}{\left(\sqrt{x} \right)}$$$ derivaatta
Aiheeseen liittyvät laskurit: Logaritmisen derivoinnin laskin, Vaiheittainen implisiittisen derivoinnin laskin
Syötteesi
Määritä $$$\frac{d}{dx} \left(\operatorname{atan}{\left(\sqrt{x} \right)}\right)$$$.
Ratkaisu
Funktio $$$\operatorname{atan}{\left(\sqrt{x} \right)}$$$ on kahden funktion $$$f{\left(u \right)} = \operatorname{atan}{\left(u \right)}$$$ ja $$$g{\left(x \right)} = \sqrt{x}$$$ yhdistelmä $$$f{\left(g{\left(x \right)} \right)}$$$.
Sovella ketjusääntöä $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$:
$${\color{red}\left(\frac{d}{dx} \left(\operatorname{atan}{\left(\sqrt{x} \right)}\right)\right)} = {\color{red}\left(\frac{d}{du} \left(\operatorname{atan}{\left(u \right)}\right) \frac{d}{dx} \left(\sqrt{x}\right)\right)}$$Arkustangentin derivaatta on $$$\frac{d}{du} \left(\operatorname{atan}{\left(u \right)}\right) = \frac{1}{u^{2} + 1}$$$:
$${\color{red}\left(\frac{d}{du} \left(\operatorname{atan}{\left(u \right)}\right)\right)} \frac{d}{dx} \left(\sqrt{x}\right) = {\color{red}\left(\frac{1}{u^{2} + 1}\right)} \frac{d}{dx} \left(\sqrt{x}\right)$$Palaa alkuperäiseen muuttujaan:
$$\frac{\frac{d}{dx} \left(\sqrt{x}\right)}{{\color{red}\left(u\right)}^{2} + 1} = \frac{\frac{d}{dx} \left(\sqrt{x}\right)}{{\color{red}\left(\sqrt{x}\right)}^{2} + 1}$$Sovella potenssisääntöä $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$, kun $$$n = \frac{1}{2}$$$:
$$\frac{{\color{red}\left(\frac{d}{dx} \left(\sqrt{x}\right)\right)}}{x + 1} = \frac{{\color{red}\left(\frac{1}{2 \sqrt{x}}\right)}}{x + 1}$$Näin ollen, $$$\frac{d}{dx} \left(\operatorname{atan}{\left(\sqrt{x} \right)}\right) = \frac{1}{2 \sqrt{x} \left(x + 1\right)}$$$.
Vastaus
$$$\frac{d}{dx} \left(\operatorname{atan}{\left(\sqrt{x} \right)}\right) = \frac{1}{2 \sqrt{x} \left(x + 1\right)}$$$A