Andra derivatan av $$$\tan^{2}{\left(x \right)}$$$

Kalkylatorn kommer att beräkna den andra derivatan av $$$\tan^{2}{\left(x \right)}$$$, med steg som visas.

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Din inmatning

Bestäm $$$\frac{d^{2}}{dx^{2}} \left(\tan^{2}{\left(x \right)}\right)$$$.

Lösning

Bestäm den första derivatan $$$\frac{d}{dx} \left(\tan^{2}{\left(x \right)}\right)$$$

Funktionen $$$\tan^{2}{\left(x \right)}$$$ är sammansättningen $$$f{\left(g{\left(x \right)} \right)}$$$ av två funktioner $$$f{\left(u \right)} = u^{2}$$$ och $$$g{\left(x \right)} = \tan{\left(x \right)}$$$.

Tillämpa kedjeregeln $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$:

$${\color{red}\left(\frac{d}{dx} \left(\tan^{2}{\left(x \right)}\right)\right)} = {\color{red}\left(\frac{d}{du} \left(u^{2}\right) \frac{d}{dx} \left(\tan{\left(x \right)}\right)\right)}$$

Tillämpa potensregeln $$$\frac{d}{du} \left(u^{n}\right) = n u^{n - 1}$$$ med $$$n = 2$$$:

$${\color{red}\left(\frac{d}{du} \left(u^{2}\right)\right)} \frac{d}{dx} \left(\tan{\left(x \right)}\right) = {\color{red}\left(2 u\right)} \frac{d}{dx} \left(\tan{\left(x \right)}\right)$$

Återgå till den ursprungliga variabeln:

$$2 {\color{red}\left(u\right)} \frac{d}{dx} \left(\tan{\left(x \right)}\right) = 2 {\color{red}\left(\tan{\left(x \right)}\right)} \frac{d}{dx} \left(\tan{\left(x \right)}\right)$$

Derivatan av tangens är $$$\frac{d}{dx} \left(\tan{\left(x \right)}\right) = \sec^{2}{\left(x \right)}$$$:

$$2 \tan{\left(x \right)} {\color{red}\left(\frac{d}{dx} \left(\tan{\left(x \right)}\right)\right)} = 2 \tan{\left(x \right)} {\color{red}\left(\sec^{2}{\left(x \right)}\right)}$$

Alltså, $$$\frac{d}{dx} \left(\tan^{2}{\left(x \right)}\right) = 2 \tan{\left(x \right)} \sec^{2}{\left(x \right)}$$$.

Därefter, $$$\frac{d^{2}}{dx^{2}} \left(\tan^{2}{\left(x \right)}\right) = \frac{d}{dx} \left(2 \tan{\left(x \right)} \sec^{2}{\left(x \right)}\right)$$$

Tillämpa konstantfaktorregeln $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$ med $$$c = 2$$$ och $$$f{\left(x \right)} = \tan{\left(x \right)} \sec^{2}{\left(x \right)}$$$:

$${\color{red}\left(\frac{d}{dx} \left(2 \tan{\left(x \right)} \sec^{2}{\left(x \right)}\right)\right)} = {\color{red}\left(2 \frac{d}{dx} \left(\tan{\left(x \right)} \sec^{2}{\left(x \right)}\right)\right)}$$

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 $$$f{\left(x \right)} = \sec^{2}{\left(x \right)}$$$ och $$$g{\left(x \right)} = \tan{\left(x \right)}$$$:

$$2 {\color{red}\left(\frac{d}{dx} \left(\tan{\left(x \right)} \sec^{2}{\left(x \right)}\right)\right)} = 2 {\color{red}\left(\frac{d}{dx} \left(\sec^{2}{\left(x \right)}\right) \tan{\left(x \right)} + \sec^{2}{\left(x \right)} \frac{d}{dx} \left(\tan{\left(x \right)}\right)\right)}$$

Derivatan av tangens är $$$\frac{d}{dx} \left(\tan{\left(x \right)}\right) = \sec^{2}{\left(x \right)}$$$:

$$2 \tan{\left(x \right)} \frac{d}{dx} \left(\sec^{2}{\left(x \right)}\right) + 2 \sec^{2}{\left(x \right)} {\color{red}\left(\frac{d}{dx} \left(\tan{\left(x \right)}\right)\right)} = 2 \tan{\left(x \right)} \frac{d}{dx} \left(\sec^{2}{\left(x \right)}\right) + 2 \sec^{2}{\left(x \right)} {\color{red}\left(\sec^{2}{\left(x \right)}\right)}$$

Funktionen $$$\sec^{2}{\left(x \right)}$$$ är sammansättningen $$$f{\left(g{\left(x \right)} \right)}$$$ av två funktioner $$$f{\left(u \right)} = u^{2}$$$ och $$$g{\left(x \right)} = \sec{\left(x \right)}$$$.

Tillämpa kedjeregeln $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$:

$$2 \tan{\left(x \right)} {\color{red}\left(\frac{d}{dx} \left(\sec^{2}{\left(x \right)}\right)\right)} + 2 \sec^{4}{\left(x \right)} = 2 \tan{\left(x \right)} {\color{red}\left(\frac{d}{du} \left(u^{2}\right) \frac{d}{dx} \left(\sec{\left(x \right)}\right)\right)} + 2 \sec^{4}{\left(x \right)}$$

Tillämpa potensregeln $$$\frac{d}{du} \left(u^{n}\right) = n u^{n - 1}$$$ med $$$n = 2$$$:

$$2 \tan{\left(x \right)} {\color{red}\left(\frac{d}{du} \left(u^{2}\right)\right)} \frac{d}{dx} \left(\sec{\left(x \right)}\right) + 2 \sec^{4}{\left(x \right)} = 2 \tan{\left(x \right)} {\color{red}\left(2 u\right)} \frac{d}{dx} \left(\sec{\left(x \right)}\right) + 2 \sec^{4}{\left(x \right)}$$

Återgå till den ursprungliga variabeln:

$$4 \tan{\left(x \right)} {\color{red}\left(u\right)} \frac{d}{dx} \left(\sec{\left(x \right)}\right) + 2 \sec^{4}{\left(x \right)} = 4 \tan{\left(x \right)} {\color{red}\left(\sec{\left(x \right)}\right)} \frac{d}{dx} \left(\sec{\left(x \right)}\right) + 2 \sec^{4}{\left(x \right)}$$

Sekantens derivata är $$$\frac{d}{dx} \left(\sec{\left(x \right)}\right) = \tan{\left(x \right)} \sec{\left(x \right)}$$$:

$$4 \tan{\left(x \right)} \sec{\left(x \right)} {\color{red}\left(\frac{d}{dx} \left(\sec{\left(x \right)}\right)\right)} + 2 \sec^{4}{\left(x \right)} = 4 \tan{\left(x \right)} \sec{\left(x \right)} {\color{red}\left(\tan{\left(x \right)} \sec{\left(x \right)}\right)} + 2 \sec^{4}{\left(x \right)}$$

Förenkla:

$$4 \tan^{2}{\left(x \right)} \sec^{2}{\left(x \right)} + 2 \sec^{4}{\left(x \right)} = \left(-4 + \frac{6}{\cos^{2}{\left(x \right)}}\right) \sec^{2}{\left(x \right)}$$

Alltså, $$$\frac{d}{dx} \left(2 \tan{\left(x \right)} \sec^{2}{\left(x \right)}\right) = \left(-4 + \frac{6}{\cos^{2}{\left(x \right)}}\right) \sec^{2}{\left(x \right)}$$$.

Således, $$$\frac{d^{2}}{dx^{2}} \left(\tan^{2}{\left(x \right)}\right) = \left(-4 + \frac{6}{\cos^{2}{\left(x \right)}}\right) \sec^{2}{\left(x \right)}$$$.

Svar

$$$\frac{d^{2}}{dx^{2}} \left(\tan^{2}{\left(x \right)}\right) = \left(-4 + \frac{6}{\cos^{2}{\left(x \right)}}\right) \sec^{2}{\left(x \right)}$$$A


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