Turunan dari $$$\tan{\left(x \right)} \sec{\left(x \right)}$$$

Terapkan aturan hasil kali $$$\frac{d}{dx} \left(f{\left(x \right)} g{\left(x \right)}\right) = \frac{d}{dx} \left(f{\left(x \right)}\right) g{\left(x \right)} + f{\left(x \right)} \frac{d}{dx} \left(g{\left(x \right)}\right)$$$ pada $$$f{\left(x \right)} = \sec{\left(x \right)}$$$ dan $$$g{\left(x \right)} = \tan{\left(x \right)}$$$:

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

Turunan fungsi sekan adalah $$$\frac{d}{dx} \left(\sec{\left(x \right)}\right) = \tan{\left(x \right)} \sec{\left(x \right)}$$$:

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

Turunan fungsi tangen adalah $$$\frac{d}{dx} \left(\tan{\left(x \right)}\right) = \sec^{2}{\left(x \right)}$$$:

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

Sederhanakan:

$$\tan^{2}{\left(x \right)} \sec{\left(x \right)} + \sec^{3}{\left(x \right)} = \left(-1 + \frac{2}{\cos^{2}{\left(x \right)}}\right) \sec{\left(x \right)}$$

Dengan demikian, $$$\frac{d}{dx} \left(\tan{\left(x \right)} \sec{\left(x \right)}\right) = \left(-1 + \frac{2}{\cos^{2}{\left(x \right)}}\right) \sec{\left(x \right)}$$$.