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Zweite Ableitung von $$$\tan^{2}{\left(x \right)}$$$
Ähnliche Rechner: Ableitungsrechner, Rechner für logarithmische Differentiation
Ihre Eingabe
Bestimme $$$\frac{d^{2}}{dx^{2}} \left(\tan^{2}{\left(x \right)}\right)$$$.
Lösung
Bestimme die erste Ableitung $$$\frac{d}{dx} \left(\tan^{2}{\left(x \right)}\right)$$$
Die Funktion $$$\tan^{2}{\left(x \right)}$$$ ist die Komposition $$$f{\left(g{\left(x \right)} \right)}$$$ der beiden Funktionen $$$f{\left(u \right)} = u^{2}$$$ und $$$g{\left(x \right)} = \tan{\left(x \right)}$$$.
Wende die Kettenregel $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$ an:
$${\color{red}\left(\frac{d}{dx} \left(\tan^{2}{\left(x \right)}\right)\right)} = {\color{red}\left(\frac{d}{du} \left(u^{2}\right) \frac{d}{dx} \left(\tan{\left(x \right)}\right)\right)}$$Wende die Potenzregel $$$\frac{d}{du} \left(u^{n}\right) = n u^{n - 1}$$$ mit $$$n = 2$$$ an:
$${\color{red}\left(\frac{d}{du} \left(u^{2}\right)\right)} \frac{d}{dx} \left(\tan{\left(x \right)}\right) = {\color{red}\left(2 u\right)} \frac{d}{dx} \left(\tan{\left(x \right)}\right)$$Zurück zur ursprünglichen Variable:
$$2 {\color{red}\left(u\right)} \frac{d}{dx} \left(\tan{\left(x \right)}\right) = 2 {\color{red}\left(\tan{\left(x \right)}\right)} \frac{d}{dx} \left(\tan{\left(x \right)}\right)$$Die Ableitung des Tangens ist $$$\frac{d}{dx} \left(\tan{\left(x \right)}\right) = \sec^{2}{\left(x \right)}$$$:
$$2 \tan{\left(x \right)} {\color{red}\left(\frac{d}{dx} \left(\tan{\left(x \right)}\right)\right)} = 2 \tan{\left(x \right)} {\color{red}\left(\sec^{2}{\left(x \right)}\right)}$$Somit gilt $$$\frac{d}{dx} \left(\tan^{2}{\left(x \right)}\right) = 2 \tan{\left(x \right)} \sec^{2}{\left(x \right)}$$$.
Als Nächstes, $$$\frac{d^{2}}{dx^{2}} \left(\tan^{2}{\left(x \right)}\right) = \frac{d}{dx} \left(2 \tan{\left(x \right)} \sec^{2}{\left(x \right)}\right)$$$
Wende die Konstantenfaktorregel $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$ mit $$$c = 2$$$ und $$$f{\left(x \right)} = \tan{\left(x \right)} \sec^{2}{\left(x \right)}$$$ an:
$${\color{red}\left(\frac{d}{dx} \left(2 \tan{\left(x \right)} \sec^{2}{\left(x \right)}\right)\right)} = {\color{red}\left(2 \frac{d}{dx} \left(\tan{\left(x \right)} \sec^{2}{\left(x \right)}\right)\right)}$$Wende die Produktregel $$$\frac{d}{dx} \left(f{\left(x \right)} g{\left(x \right)}\right) = \frac{d}{dx} \left(f{\left(x \right)}\right) g{\left(x \right)} + f{\left(x \right)} \frac{d}{dx} \left(g{\left(x \right)}\right)$$$ mit $$$f{\left(x \right)} = \sec^{2}{\left(x \right)}$$$ und $$$g{\left(x \right)} = \tan{\left(x \right)}$$$ an:
$$2 {\color{red}\left(\frac{d}{dx} \left(\tan{\left(x \right)} \sec^{2}{\left(x \right)}\right)\right)} = 2 {\color{red}\left(\frac{d}{dx} \left(\sec^{2}{\left(x \right)}\right) \tan{\left(x \right)} + \sec^{2}{\left(x \right)} \frac{d}{dx} \left(\tan{\left(x \right)}\right)\right)}$$Die Ableitung des Tangens ist $$$\frac{d}{dx} \left(\tan{\left(x \right)}\right) = \sec^{2}{\left(x \right)}$$$:
$$2 \tan{\left(x \right)} \frac{d}{dx} \left(\sec^{2}{\left(x \right)}\right) + 2 \sec^{2}{\left(x \right)} {\color{red}\left(\frac{d}{dx} \left(\tan{\left(x \right)}\right)\right)} = 2 \tan{\left(x \right)} \frac{d}{dx} \left(\sec^{2}{\left(x \right)}\right) + 2 \sec^{2}{\left(x \right)} {\color{red}\left(\sec^{2}{\left(x \right)}\right)}$$Die Funktion $$$\sec^{2}{\left(x \right)}$$$ ist die Komposition $$$f{\left(g{\left(x \right)} \right)}$$$ der beiden Funktionen $$$f{\left(u \right)} = u^{2}$$$ und $$$g{\left(x \right)} = \sec{\left(x \right)}$$$.
Wende die Kettenregel $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$ an:
$$2 \tan{\left(x \right)} {\color{red}\left(\frac{d}{dx} \left(\sec^{2}{\left(x \right)}\right)\right)} + 2 \sec^{4}{\left(x \right)} = 2 \tan{\left(x \right)} {\color{red}\left(\frac{d}{du} \left(u^{2}\right) \frac{d}{dx} \left(\sec{\left(x \right)}\right)\right)} + 2 \sec^{4}{\left(x \right)}$$Wende die Potenzregel $$$\frac{d}{du} \left(u^{n}\right) = n u^{n - 1}$$$ mit $$$n = 2$$$ an:
$$2 \tan{\left(x \right)} {\color{red}\left(\frac{d}{du} \left(u^{2}\right)\right)} \frac{d}{dx} \left(\sec{\left(x \right)}\right) + 2 \sec^{4}{\left(x \right)} = 2 \tan{\left(x \right)} {\color{red}\left(2 u\right)} \frac{d}{dx} \left(\sec{\left(x \right)}\right) + 2 \sec^{4}{\left(x \right)}$$Zurück zur ursprünglichen Variable:
$$4 \tan{\left(x \right)} {\color{red}\left(u\right)} \frac{d}{dx} \left(\sec{\left(x \right)}\right) + 2 \sec^{4}{\left(x \right)} = 4 \tan{\left(x \right)} {\color{red}\left(\sec{\left(x \right)}\right)} \frac{d}{dx} \left(\sec{\left(x \right)}\right) + 2 \sec^{4}{\left(x \right)}$$Die Ableitung des Sekans ist $$$\frac{d}{dx} \left(\sec{\left(x \right)}\right) = \tan{\left(x \right)} \sec{\left(x \right)}$$$:
$$4 \tan{\left(x \right)} \sec{\left(x \right)} {\color{red}\left(\frac{d}{dx} \left(\sec{\left(x \right)}\right)\right)} + 2 \sec^{4}{\left(x \right)} = 4 \tan{\left(x \right)} \sec{\left(x \right)} {\color{red}\left(\tan{\left(x \right)} \sec{\left(x \right)}\right)} + 2 \sec^{4}{\left(x \right)}$$Vereinfachen:
$$4 \tan^{2}{\left(x \right)} \sec^{2}{\left(x \right)} + 2 \sec^{4}{\left(x \right)} = \left(-4 + \frac{6}{\cos^{2}{\left(x \right)}}\right) \sec^{2}{\left(x \right)}$$Somit gilt $$$\frac{d}{dx} \left(2 \tan{\left(x \right)} \sec^{2}{\left(x \right)}\right) = \left(-4 + \frac{6}{\cos^{2}{\left(x \right)}}\right) \sec^{2}{\left(x \right)}$$$.
Daher $$$\frac{d^{2}}{dx^{2}} \left(\tan^{2}{\left(x \right)}\right) = \left(-4 + \frac{6}{\cos^{2}{\left(x \right)}}\right) \sec^{2}{\left(x \right)}$$$.
Antwort
$$$\frac{d^{2}}{dx^{2}} \left(\tan^{2}{\left(x \right)}\right) = \left(-4 + \frac{6}{\cos^{2}{\left(x \right)}}\right) \sec^{2}{\left(x \right)}$$$A