$$$\tan^{2}{\left(x \right)}$$$'nin ikinci türevi
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Girdiniz
Bulun: $$$\frac{d^{2}}{dx^{2}} \left(\tan^{2}{\left(x \right)}\right)$$$.
Çözüm
Birinci türevi bulun $$$\frac{d}{dx} \left(\tan^{2}{\left(x \right)}\right)$$$
$$$\tan^{2}{\left(x \right)}$$$ fonksiyonu, iki fonksiyon $$$f{\left(u \right)} = u^{2}$$$ ve $$$g{\left(x \right)} = \tan{\left(x \right)}$$$'nin $$$f{\left(g{\left(x \right)} \right)}$$$ bileşimidir.
Zincir kuralını $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$ uygulayın:
$${\color{red}\left(\frac{d}{dx} \left(\tan^{2}{\left(x \right)}\right)\right)} = {\color{red}\left(\frac{d}{du} \left(u^{2}\right) \frac{d}{dx} \left(\tan{\left(x \right)}\right)\right)}$$$$$\frac{d}{du} \left(u^{n}\right) = n u^{n - 1}$$$ şeklindeki kuvvet kuralını $$$n = 2$$$ ile uygula:
$${\color{red}\left(\frac{d}{du} \left(u^{2}\right)\right)} \frac{d}{dx} \left(\tan{\left(x \right)}\right) = {\color{red}\left(2 u\right)} \frac{d}{dx} \left(\tan{\left(x \right)}\right)$$Eski değişkene geri dön:
$$2 {\color{red}\left(u\right)} \frac{d}{dx} \left(\tan{\left(x \right)}\right) = 2 {\color{red}\left(\tan{\left(x \right)}\right)} \frac{d}{dx} \left(\tan{\left(x \right)}\right)$$Tanjantın türevi $$$\frac{d}{dx} \left(\tan{\left(x \right)}\right) = \sec^{2}{\left(x \right)}$$$:
$$2 \tan{\left(x \right)} {\color{red}\left(\frac{d}{dx} \left(\tan{\left(x \right)}\right)\right)} = 2 \tan{\left(x \right)} {\color{red}\left(\sec^{2}{\left(x \right)}\right)}$$Dolayısıyla, $$$\frac{d}{dx} \left(\tan^{2}{\left(x \right)}\right) = 2 \tan{\left(x \right)} \sec^{2}{\left(x \right)}$$$.
Ardından, $$$\frac{d^{2}}{dx^{2}} \left(\tan^{2}{\left(x \right)}\right) = \frac{d}{dx} \left(2 \tan{\left(x \right)} \sec^{2}{\left(x \right)}\right)$$$
Sabit çarpan kuralını $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$ $$$c = 2$$$ ve $$$f{\left(x \right)} = \tan{\left(x \right)} \sec^{2}{\left(x \right)}$$$ ile uygula:
$${\color{red}\left(\frac{d}{dx} \left(2 \tan{\left(x \right)} \sec^{2}{\left(x \right)}\right)\right)} = {\color{red}\left(2 \frac{d}{dx} \left(\tan{\left(x \right)} \sec^{2}{\left(x \right)}\right)\right)}$$Çarpım kuralını $$$f{\left(x \right)} = \sec^{2}{\left(x \right)}$$$ ve $$$g{\left(x \right)} = \tan{\left(x \right)}$$$ ile $$$\frac{d}{dx} \left(f{\left(x \right)} g{\left(x \right)}\right) = \frac{d}{dx} \left(f{\left(x \right)}\right) g{\left(x \right)} + f{\left(x \right)} \frac{d}{dx} \left(g{\left(x \right)}\right)$$$ kullanarak uygulayın:
$$2 {\color{red}\left(\frac{d}{dx} \left(\tan{\left(x \right)} \sec^{2}{\left(x \right)}\right)\right)} = 2 {\color{red}\left(\frac{d}{dx} \left(\sec^{2}{\left(x \right)}\right) \tan{\left(x \right)} + \sec^{2}{\left(x \right)} \frac{d}{dx} \left(\tan{\left(x \right)}\right)\right)}$$Tanjantın türevi $$$\frac{d}{dx} \left(\tan{\left(x \right)}\right) = \sec^{2}{\left(x \right)}$$$:
$$2 \tan{\left(x \right)} \frac{d}{dx} \left(\sec^{2}{\left(x \right)}\right) + 2 \sec^{2}{\left(x \right)} {\color{red}\left(\frac{d}{dx} \left(\tan{\left(x \right)}\right)\right)} = 2 \tan{\left(x \right)} \frac{d}{dx} \left(\sec^{2}{\left(x \right)}\right) + 2 \sec^{2}{\left(x \right)} {\color{red}\left(\sec^{2}{\left(x \right)}\right)}$$$$$\sec^{2}{\left(x \right)}$$$ fonksiyonu, iki fonksiyon $$$f{\left(u \right)} = u^{2}$$$ ve $$$g{\left(x \right)} = \sec{\left(x \right)}$$$'nin $$$f{\left(g{\left(x \right)} \right)}$$$ bileşimidir.
Zincir kuralını $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$ uygulayın:
$$2 \tan{\left(x \right)} {\color{red}\left(\frac{d}{dx} \left(\sec^{2}{\left(x \right)}\right)\right)} + 2 \sec^{4}{\left(x \right)} = 2 \tan{\left(x \right)} {\color{red}\left(\frac{d}{du} \left(u^{2}\right) \frac{d}{dx} \left(\sec{\left(x \right)}\right)\right)} + 2 \sec^{4}{\left(x \right)}$$$$$\frac{d}{du} \left(u^{n}\right) = n u^{n - 1}$$$ şeklindeki kuvvet kuralını $$$n = 2$$$ ile uygula:
$$2 \tan{\left(x \right)} {\color{red}\left(\frac{d}{du} \left(u^{2}\right)\right)} \frac{d}{dx} \left(\sec{\left(x \right)}\right) + 2 \sec^{4}{\left(x \right)} = 2 \tan{\left(x \right)} {\color{red}\left(2 u\right)} \frac{d}{dx} \left(\sec{\left(x \right)}\right) + 2 \sec^{4}{\left(x \right)}$$Eski değişkene geri dön:
$$4 \tan{\left(x \right)} {\color{red}\left(u\right)} \frac{d}{dx} \left(\sec{\left(x \right)}\right) + 2 \sec^{4}{\left(x \right)} = 4 \tan{\left(x \right)} {\color{red}\left(\sec{\left(x \right)}\right)} \frac{d}{dx} \left(\sec{\left(x \right)}\right) + 2 \sec^{4}{\left(x \right)}$$Sekantın türevi $$$\frac{d}{dx} \left(\sec{\left(x \right)}\right) = \tan{\left(x \right)} \sec{\left(x \right)}$$$:
$$4 \tan{\left(x \right)} \sec{\left(x \right)} {\color{red}\left(\frac{d}{dx} \left(\sec{\left(x \right)}\right)\right)} + 2 \sec^{4}{\left(x \right)} = 4 \tan{\left(x \right)} \sec{\left(x \right)} {\color{red}\left(\tan{\left(x \right)} \sec{\left(x \right)}\right)} + 2 \sec^{4}{\left(x \right)}$$Sadeleştirin:
$$4 \tan^{2}{\left(x \right)} \sec^{2}{\left(x \right)} + 2 \sec^{4}{\left(x \right)} = \left(-4 + \frac{6}{\cos^{2}{\left(x \right)}}\right) \sec^{2}{\left(x \right)}$$Dolayısıyla, $$$\frac{d}{dx} \left(2 \tan{\left(x \right)} \sec^{2}{\left(x \right)}\right) = \left(-4 + \frac{6}{\cos^{2}{\left(x \right)}}\right) \sec^{2}{\left(x \right)}$$$.
Dolayısıyla, $$$\frac{d^{2}}{dx^{2}} \left(\tan^{2}{\left(x \right)}\right) = \left(-4 + \frac{6}{\cos^{2}{\left(x \right)}}\right) \sec^{2}{\left(x \right)}$$$.
Cevap
$$$\frac{d^{2}}{dx^{2}} \left(\tan^{2}{\left(x \right)}\right) = \left(-4 + \frac{6}{\cos^{2}{\left(x \right)}}\right) \sec^{2}{\left(x \right)}$$$A